Here is a lower bound that improves Joel's by a bit. If the reflection property of the post holds at $\kappa$ then there is some measurable $\lambda<\kappa$ with $$V_\lambda\models``\text{there is a proper class of extendible cardinals}".$$ In fact, we can have $\lambda$ itself somewhat extendible if we want to. Let $A, B$ be as in Joel's answer, so for any $b\in B$ there is an elementary $j_b:A\rightarrow B$ with $b\in\mathrm{ran}(j)$.

We now define an increasing sequence $\langle \bar\kappa_\beta\mid\beta<\alpha\rangle$ and only stop when we hit a limit $\alpha$ so that $\sup_{\beta<\alpha}\bar\kappa_\beta$ is measurable. By recursion, we let $$\bar\kappa_\beta:=\min\{\mathrm{crit}(j_{\{b, \langle \bar\kappa_\gamma\mid\gamma<\beta\rangle\}})\mid b\in B\}.$$ We have to show that we hit the stopping $\alpha$ before the sequence is no longer increasing. So suppose instead that we find some $\beta<\xi$ with $\bar\kappa_\xi\leq\bar\kappa_\beta$ first and let $(\xi,\beta)$ be lexicographically least so. Now we can find $j:A\rightarrow B$ with $\mathrm{crit}(j)=\bar\kappa_\xi$ and $\langle\bar\kappa_\gamma\mid\gamma<\xi\rangle$ in the range of $j$, say its preimage is $\langle \bar\lambda_\gamma\mid\gamma<\bar\xi\rangle$. If defined, we have $\bar\kappa_\xi\leq\bar\kappa_{\bar\kappa_\xi}$, so by minimality of $\beta$ necessarily $\beta\leq\bar\kappa_\xi$. If we had $\beta<\bar\kappa_\xi$ then $j(\langle\bar\lambda_\gamma\mid\gamma<\beta\rangle)=\langle \bar\kappa_\gamma\mid\gamma<\beta\rangle$ is in the range of $j$, so that $\bar\kappa_\beta=\bar\kappa_\xi$ by definition of $\bar\kappa_\beta$. But $\bar\kappa_\beta=j(\bar\lambda_\beta)$ is in the range of $j$, yet $\bar\kappa_\beta=\mathrm{crit}(j)$, contradiction. We are left with $\bar\kappa_\xi=\beta$. Joel's argument shows that $\beta$ must be measurable: We only need to see $\mathcal P(\beta)\subseteq A$. Now for $X\subseteq\beta$, $X$ is in the range of $j_{\{X,\langle \bar\kappa_\gamma\mid\gamma<\xi\rangle\}}$ and this map has critical point $\geq \beta$. This can only happen if already $X\in A$. We have $\bar\kappa_\gamma<\bar\kappa_\xi=\beta$ for $\gamma<\beta$ by minimality of $\beta$, so that $\sup_{\gamma<\beta}\bar\kappa_\gamma=\beta$ should have been our stopping point, contradiction!

So we set $\lambda=\sup_{\beta<\alpha}\kappa_\beta$ (actually $\lambda=\alpha)$ which is measurable by construction. We have that $\langle \bar\kappa_\beta\mid\beta<\lambda\rangle$ is an increasing cofinal sequence in $\lambda$ and we will show that unboundedly many of those are extendible in $V_\lambda$. Once again, $V_\lambda\subseteq A$ by applying Joel's argument to all $\bar\kappa_\beta$ for $\beta<\lambda$.

So any $\bar\kappa_\beta$ is $\lambda$-extendible, i.e. there is some $\nu_\beta$ and $k_\beta\colon V_\lambda\rightarrow V_{\nu_\beta}$ with critical point $\bar\kappa_\beta$. By Dushnik-Miller, there is some $H\in[\lambda]^\lambda$ so that $\nu_\beta\leq\nu_\gamma$ for every $\beta<\gamma$ both in $H$.

We are done once we show that $\bar\kappa_\beta$ is extendible in $V_\lambda$ for $\beta\in H$, since those $\bar\kappa_\beta$ are cofinal in $\lambda$. It is enough to show that if $\beta<\gamma$ are both in $H$ then $\bar\kappa_\beta$ is ${<}\bar\kappa_\gamma$-extendible in $V_\lambda$. Let $$\mu:=k_\beta(\bar\kappa_\gamma)<\nu_\beta\leq\nu_\gamma.$$ Then $k_\beta\upharpoonright V_{\bar\kappa_\gamma}:V_{\bar\kappa_\gamma}\rightarrow V_\mu$ is elementary, has critical point $\bar\kappa_\beta$ and is in $V_{\nu_\gamma}$. By elementarity of $k_\gamma$, we must have $V_\lambda\models``\bar\kappa_\beta\text{ is $\delta$-extendible}"$ for each $\delta<\bar\kappa_\gamma$.

Here is a lower bound that improves Joel's by a bit. If the reflection property of the post holds at $\kappa$ then there is some measurable $\lambda<\kappa$ with $$V_\lambda\models``\text{there is a proper class of extendible cardinals}".$$ In fact, we can have $\lambda$ itself somewhat extendible if we want to. Let $A, B$ be as in Joel's answer, so for any $b\in B$ there is an elementary $j_b:A\rightarrow B$ with $b\in\mathrm{ran}(j_b)$ (and say we pick $j_b$ with minimal possible critical point).

We now define an increasing sequence $\langle \bar\kappa_\beta\mid\beta<\alpha\rangle$ and only stop when we hit a limit $\alpha$ so that $\sup_{\beta<\alpha}\bar\kappa_\beta$ is measurable. By recursion, we let $$\bar\kappa_\beta:=\min\{\mathrm{crit}(j_{\{b, \langle \bar\kappa_\gamma\mid\gamma<\beta\rangle\}})\mid b\in B\}.$$ We have to show that we hit the stopping $\alpha$ before the sequence is no longer increasing. So suppose instead that we find some $\beta<\xi$ with $\bar\kappa_\xi\leq\bar\kappa_\beta$ first and let $(\xi,\beta)$ be lexicographically least so. Now we can find $j:A\rightarrow B$ with $\mathrm{crit}(j)=\bar\kappa_\xi$ and $\langle\bar\kappa_\gamma\mid\gamma<\xi\rangle$ in the range of $j$, say its preimage is $\langle \bar\lambda_\gamma\mid\gamma<\bar\xi\rangle$. If defined, we have $\bar\kappa_\xi\leq\bar\kappa_{\bar\kappa_\xi}$, so by minimality of $\beta$ necessarily $\beta\leq\bar\kappa_\xi$. If we had $\beta<\bar\kappa_\xi$ then $j(\langle\bar\lambda_\gamma\mid\gamma<\beta\rangle)=\langle \bar\kappa_\gamma\mid\gamma<\beta\rangle$ is in the range of $j$, so that $\bar\kappa_\beta=\bar\kappa_\xi$ by definition of $\bar\kappa_\beta$. But $\bar\kappa_\beta=j(\bar\lambda_\beta)$ is in the range of $j$, yet $\bar\kappa_\beta=\mathrm{crit}(j)$, contradiction. We are left with $\bar\kappa_\xi=\beta$. Joel's argument shows that $\beta$ must be measurable: We only need to see $\mathcal P(\beta)\subseteq A$. Now for $X\subseteq\beta$, $X$ is in the range of $j_{\{X,\langle \bar\kappa_\gamma\mid\gamma<\xi\rangle\}}$ and this map has critical point $\geq \beta$. This can only happen if already $X\in A$. We have $\bar\kappa_\gamma<\bar\kappa_\xi=\beta$ for $\gamma<\beta$ by minimality of $\beta$, so that $\sup_{\gamma<\beta}\bar\kappa_\gamma=\beta$ should have been our stopping point, contradiction!

So we set $\lambda=\sup_{\beta<\alpha}\kappa_\beta$ (actually $\lambda=\alpha)$ which is measurable by construction. We have that $\langle \bar\kappa_\beta\mid\beta<\lambda\rangle$ is an increasing cofinal sequence in $\lambda$ and we will show that unboundedly many of those are extendible in $V_\lambda$. Once again, $V_\lambda\subseteq A$ by applying Joel's argument to all $\bar\kappa_\beta$ for $\beta<\lambda$.

So any $\bar\kappa_\beta$ is $\lambda$-extendible, i.e. there is some $\nu_\beta$ and $k_\beta\colon V_\lambda\rightarrow V_{\nu_\beta}$ with critical point $\bar\kappa_\beta$. By Dushnik-Miller, there is some $H\in[\lambda]^\lambda$ so that $\nu_\beta\leq\nu_\gamma$ for every $\beta<\gamma$ both in $H$.

We are done once we show that $\bar\kappa_\beta$ is extendible in $V_\lambda$ for $\beta\in H$, since those $\bar\kappa_\beta$ are cofinal in $\lambda$. It is enough to show that if $\beta<\gamma$ are both in $H$ then $\bar\kappa_\beta$ is ${<}\bar\kappa_\gamma$-extendible in $V_\lambda$. Let $$\mu:=k_\beta(\bar\kappa_\gamma)<\nu_\beta\leq\nu_\gamma.$$ Then $k_\beta\upharpoonright V_{\bar\kappa_\gamma}:V_{\bar\kappa_\gamma}\rightarrow V_\mu$ is elementary, has critical point $\bar\kappa_\beta$ and is in $V_{\nu_\gamma}$. By elementarity of $k_\gamma$, we must have $V_\lambda\models``\bar\kappa_\beta\text{ is $\delta$-extendible}"$ for each $\delta<\bar\kappa_\gamma$.

Here is a lower bound that improves Joel's by a bit. If the reflection property of the post holds at $\kappa$ then there is some measurable $\lambda<\kappa$ with $$V_\lambda\models``\text{there is a proper class of extendible cardinals}".$$ In fact, we can have $\lambda$ itself somewhat extendible if we want to. Let $A, B$ be as in Joel's answer, so for any $b\in B$ there is an elementary $j_b:A\rightarrow B$ with $b\in\mathrm{ran}(j)$.

We now define an increasing sequence $\langle \bar\kappa_\beta\mid\beta<\alpha\rangle$ and only stop when $\sup_{\beta<\alpha}\bar\kappa_\beta$ is measurable. By recursion, we let $$\bar\kappa_\beta:=\min\{\mathrm{crit}(j_{\{b, \langle \bar\kappa_\gamma\mid\gamma<\beta\rangle\}})\mid b\in B\}.$$ We have to show that we hit the stopping $\alpha$ before the sequence is no longer increasing. So suppose instead that we find some $\beta<\xi$ with $\bar\kappa_\xi\leq\bar\kappa_\beta$ first and let $(\xi,\beta)$ be lexicographically least so. Now we can find $j:A\rightarrow B$ with $\mathrm{crit}(j)=\bar\kappa_\xi$ and $\langle\bar\kappa_\gamma\mid\gamma<\xi\rangle$ in the range of $j$, say its preimage is $\langle \bar\lambda_\gamma\mid\gamma<\bar\xi\rangle$. If defined, we have $\bar\kappa_\xi\leq\bar\kappa_{\bar\kappa_\xi}$, so by minimality of $\beta$ necessarily $\beta\leq\bar\kappa_\xi$. If we had $\beta<\bar\kappa_\xi$ then $j(\langle\bar\lambda_\gamma\mid\gamma<\beta\rangle)=\langle \bar\kappa_\gamma\mid\gamma<\beta\rangle$ is in the range of $j$, so that $\bar\kappa_\beta=\bar\kappa_\xi$ by definition of $\bar\kappa_\beta$. But $\bar\kappa_\beta=j(\bar\lambda_\beta)$ is in the range of $j$, yet $\bar\kappa_\beta=\mathrm{crit}(j)$, contradiction. We are left with $\bar\kappa_\xi=\beta$. Joel's argument shows that $\beta$ must be measurable: We only need to see $\mathcal P(\beta)\subseteq A$. Now for $X\subseteq\beta$, $X$ is in the range of $j_{\{X,\langle \bar\kappa_\gamma\mid\gamma<\xi\rangle\}}$ and this map has critical point $\geq \beta$. This can only happen if already $X\in A$. We have $\bar\kappa_\gamma<\bar\kappa_\xi=\beta$ for $\gamma<\beta$ by minimality of $\beta$, so that $\sup_{\gamma<\beta}\bar\kappa_\gamma=\beta$ should have been our stopping point, contradiction!

So we set $\lambda=\sup_{\beta<\alpha}\kappa_\beta$ (actually $\lambda=\alpha)$ which is measurable by construction. We have that $\langle \bar\kappa_\beta\mid\beta<\lambda\rangle$ is an increasing cofinal sequence in $\lambda$ and we will show that unboundedly many of those are extendible in $V_\lambda$. Once again, $V_\lambda\subseteq A$ by applying Joel's argument to all $\bar\kappa_\beta$ for $\beta<\lambda$.

So any $\bar\kappa_\beta$ is $\lambda$-extendible, i.e. there is some $\nu_\beta$ and $k_\beta\colon V_\lambda\rightarrow V_{\nu_\beta}$ with critical point $\bar\kappa_\beta$. By Dushnik-Miller, there is some $H\in[\lambda]^\lambda$ so that $\nu_\beta\leq\nu_\gamma$ for every $\beta<\gamma$ both in $H$.

We are done once we show that $\bar\kappa_\beta$ is extendible in $V_\lambda$ for $\beta\in H$, since those $\bar\kappa_\beta$ are cofinal in $\lambda$. It is enough to show that if $\beta<\gamma$ are both in $H$ then $\bar\kappa_\beta$ is ${<}\bar\kappa_\gamma$-extendible in $V_\lambda$. Let $$\mu:=k_\beta(\bar\kappa_\gamma)<\nu_\beta\leq\nu_\gamma.$$ Then $k_\beta\upharpoonright V_{\bar\kappa_\gamma}:V_{\bar\kappa_\gamma}\rightarrow V_\mu$ is elementary, has critical point $\bar\kappa_\beta$ and is in $V_{\nu_\gamma}$. By elementarity of $k_\gamma$, we must have $V_\lambda\models``\bar\kappa_\beta\text{ is $\delta$-extendible}"$ for each $\delta<\bar\kappa_\gamma$.

Here is a lower bound that improves Joel's by a bit. If the reflection property of the post holds at $\kappa$ then there is some measurable $\lambda<\kappa$ with $$V_\lambda\models``\text{there is a proper class of extendible cardinals}".$$ In fact, we can have $\lambda$ itself somewhat extendible if we want to. Let $A, B$ be as in Joel's answer, so for any $b\in B$ there is an elementary $j_b:A\rightarrow B$ with $b\in\mathrm{ran}(j)$.

We now define an increasing sequence $\langle \bar\kappa_\beta\mid\beta<\alpha\rangle$ and only stop when we hit a limit $\alpha$ so that $\sup_{\beta<\alpha}\bar\kappa_\beta$ is measurable. By recursion, we let $$\bar\kappa_\beta:=\min\{\mathrm{crit}(j_{\{b, \langle \bar\kappa_\gamma\mid\gamma<\beta\rangle\}})\mid b\in B\}.$$ We have to show that we hit the stopping $\alpha$ before the sequence is no longer increasing. So suppose instead that we find some $\beta<\xi$ with $\bar\kappa_\xi\leq\bar\kappa_\beta$ first and let $(\xi,\beta)$ be lexicographically least so. Now we can find $j:A\rightarrow B$ with $\mathrm{crit}(j)=\bar\kappa_\xi$ and $\langle\bar\kappa_\gamma\mid\gamma<\xi\rangle$ in the range of $j$, say its preimage is $\langle \bar\lambda_\gamma\mid\gamma<\bar\xi\rangle$. If defined, we have $\bar\kappa_\xi\leq\bar\kappa_{\bar\kappa_\xi}$, so by minimality of $\beta$ necessarily $\beta\leq\bar\kappa_\xi$. If we had $\beta<\bar\kappa_\xi$ then $j(\langle\bar\lambda_\gamma\mid\gamma<\beta\rangle)=\langle \bar\kappa_\gamma\mid\gamma<\beta\rangle$ is in the range of $j$, so that $\bar\kappa_\beta=\bar\kappa_\xi$ by definition of $\bar\kappa_\beta$. But $\bar\kappa_\beta=j(\bar\lambda_\beta)$ is in the range of $j$, yet $\bar\kappa_\beta=\mathrm{crit}(j)$, contradiction. We are left with $\bar\kappa_\xi=\beta$. Joel's argument shows that $\beta$ must be measurable: We only need to see $\mathcal P(\beta)\subseteq A$. Now for $X\subseteq\beta$, $X$ is in the range of $j_{\{X,\langle \bar\kappa_\gamma\mid\gamma<\xi\rangle\}}$ and this map has critical point $\geq \beta$. This can only happen if already $X\in A$. We have $\bar\kappa_\gamma<\bar\kappa_\xi=\beta$ for $\gamma<\beta$ by minimality of $\beta$, so that $\sup_{\gamma<\beta}\bar\kappa_\gamma=\beta$ should have been our stopping point, contradiction!

So we set $\lambda=\sup_{\beta<\alpha}\kappa_\beta$ (actually $\lambda=\alpha)$ which is measurable by construction. We have that $\langle \bar\kappa_\beta\mid\beta<\lambda\rangle$ is an increasing cofinal sequence in $\lambda$ and we will show that unboundedly many of those are extendible in $V_\lambda$. Once again, $V_\lambda\subseteq A$ by applying Joel's argument to all $\bar\kappa_\beta$ for $\beta<\lambda$.

So any $\bar\kappa_\beta$ is $\lambda$-extendible, i.e. there is some $\nu_\beta$ and $k_\beta\colon V_\lambda\rightarrow V_{\nu_\beta}$ with critical point $\bar\kappa_\beta$. By Dushnik-Miller, there is some $H\in[\lambda]^\lambda$ so that $\nu_\beta\leq\nu_\gamma$ for every $\beta<\gamma$ both in $H$.

We are done once we show that $\bar\kappa_\beta$ is extendible in $V_\lambda$ for $\beta\in H$, since those $\bar\kappa_\beta$ are cofinal in $\lambda$. It is enough to show that if $\beta<\gamma$ are both in $H$ then $\bar\kappa_\beta$ is ${<}\bar\kappa_\gamma$-extendible in $V_\lambda$. Let $$\mu:=k_\beta(\bar\kappa_\gamma)<\nu_\beta\leq\nu_\gamma.$$ Then $k_\beta\upharpoonright V_{\bar\kappa_\gamma}:V_{\bar\kappa_\gamma}\rightarrow V_\mu$ is elementary, has critical point $\bar\kappa_\beta$ and is in $V_{\nu_\gamma}$. By elementarity of $k_\gamma$, we must have $V_\lambda\models``\bar\kappa_\beta\text{ is $\delta$-extendible}"$ for each $\delta<\bar\kappa_\gamma$.