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Bounty Ended with 100 reputation awarded by Noah Schweber
Worked out precise value of $\alpha$.
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Farmer S
Farmer S
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(EDIT: Now edited to compute the precise value of $\alpha$.)

@AsafKaragila already answered the question, but this is addressinganswering the follow-up question od @Reflecting_Ordinal in the comments to Asaf's question.

(i) $\kappa_0\leq\alpha\leq\kappa_1+\beta$$\alpha=\kappa_0+\beta$,

(i') there is a coded version of some such $M$ in $L_{\kappa_0+1}$; that is, there is a witnessing $M$ such that if $t$ is the $\Sigma_1$ theory of $M$ in parameters in $\beta\cup\{z\}$ where $z=(0^\#)^M$, and $t'$ is the theory which results by replacing $z$ with some constant symbol in $V_\omega\backslash\omega$, then $t'\in L_{\kappa_0+1}$,

(iii) $L_{\kappa_0}$ projects to $\omega$; i.e. $L_{\kappa_0+1}\cap\mathcal{P}(\omega)\not\subseteq L_{\kappa_0}$.

(I'm not sure where exactly $\alpha$ sits in the interval $[\kappa_0,\kappa_1+\beta]$.)

Note now that it follows that if we minimize on $\alpha$ first, giving say $\alpha'$, with some such $M\in L_{\alpha'+1}$, and then take the least witness $\beta'$ for $\mathrm{OR}^M$, then $\alpha'=\alpha$ and $\beta'=\beta$. For otherwise $\alpha'<\alpha$ but $\beta<\beta'$. But certainly $L_{\beta'}\models\mathrm{ZF}$, but the first ZF level beyond $L_\beta$ is well past $L_{\kappa_1+\beta}$$L_{\kappa_0+\beta}$, so $\alpha<\beta'\leq\alpha'$, contradiction.

So we have $\gamma=\kappa_0$. This gives $\alpha\leq\kappa_1+\beta$, completingwhere $\kappa_1$ is the proofleast admissible $>\kappa_0$. But this doesn't suffice for (i).

Remark: Note that it also follows that $\kappa_0\notin\mathrm{wfp}(D)$ (otherwise we get surjection of $\omega$ onto $\beta$ definable over $M$), so $\kappa_0=\mathrm{wfp}(D)$.

Now for (i'): Because $L_{\kappa_0}$ projects to $\omega$, there is a surjection $\pi:\omega\to L_{\kappa_0}$ which is definable over $L_{\kappa_0}$. Therefore, we can fix a surjection $\sigma:\omega\to \mathscr{D}$, where $\mathscr{D}=$ the set of all dense subsets of $\mathbb{P}=\mathrm{Coll}(\omega,\beta)$ which are boldface-$\Sigma_2^{L_{\kappa_0}}$-definable, and such that $\sigma$ is definable over $L_{\kappa_0}$. Let $G$ be the filter $\subseteq\mathbb{P}$ which results by meeting all the sets in $\mathscr{D}$, one by one in the usual way, using the surjection $\sigma$. So $G\subseteq L_\beta$ and $G$ is definable over $L_{\kappa_0}$, so $G\in L_{\kappa_0+1}$. Now because $G$ meets enough dense sets, $L_{\kappa_0}[G]$ is also admissible, and contains a real $x$ coding $\beta$ (just given by $G$), such that letting $\tau:\omega\to\beta$ be the corresponding surjection, then $\tau$ is in $L_{\kappa_0}[G]$. Therefore like before, there is a model $M$ of the desired form coded by a real $m$, such that $m$ is definable from parameters over $L_{\kappa_0}[G]$, and such that the coding of ordinals ${<\beta}$ given by $m$ agrees with $\tau$; in fact, we can take $m$ to be $\Sigma_1\wedge\Pi_1$-definable from parameters over $L_{\kappa_0}[G]$, considering the complexity of the left-most-branch which yields $m$. But using $G$ and the $(\Sigma_1,\Pi_1)$-forcing relation over $L_{\kappa_0}$ (for $\mathbb{P}$), and names for the relevant parameters, we can define $m$ over $L_{\kappa_0}$, and also the theory $t'$ as described in (i'), associated to the model $M$. This gives (i').

Finally for (i), we get $\alpha\leq\kappa_0+\beta$ by (i'), as it takes at most $\beta$ steps to transitivize $m$. But actually, it takes exactly $\beta$ steps, because $(0^\#)^M$ is a real which is in $M$ which first appers in $L_{\kappa_0+1}$, and so note that the sets of the form $\ldots\{\{(0^\#)^M\}\}\ldots$ (with ${<\beta}$-many nested pairs of brackets, wellfounded), which are all in $M$, take $\beta$ stages of construction after $L_{\kappa_0+1}$ to produce. Therefore $\alpha=\kappa_0+\beta$. (But really, the more important ordinal here is $\kappa_0$.)

@AsafKaragila already answered the question, but this is addressing the follow-up question od @Reflecting_Ordinal in the comments to Asaf's question.

(i) $\kappa_0\leq\alpha\leq\kappa_1+\beta$,

(iii) $L_{\kappa_0}$ projects to $\omega$; i.e. $L_{\kappa_0+1}\cap\mathcal{P}(\omega)\not\subseteq L_{\kappa_0}$.

(I'm not sure where exactly $\alpha$ sits in the interval $[\kappa_0,\kappa_1+\beta]$.)

Note now that it follows that if we minimize on $\alpha$ first, giving say $\alpha'$, with some such $M\in L_{\alpha'+1}$, and then take the least witness $\beta'$ for $\mathrm{OR}^M$, then $\alpha'=\alpha$ and $\beta'=\beta$. For otherwise $\alpha'<\alpha$ but $\beta<\beta'$. But certainly $L_{\beta'}\models\mathrm{ZF}$, but the first ZF level beyond $L_\beta$ is well past $L_{\kappa_1+\beta}$, so $\alpha<\beta'\leq\alpha'$, contradiction.

So we have $\gamma=\kappa_0$, completing the proof.

Remark: Note that it also follows that $\kappa_0\notin\mathrm{wfp}(D)$ (otherwise we get surjection of $\omega$ onto $\beta$ definable over $M$), so $\kappa_0=\mathrm{wfp}(D)$.

(EDIT: Now edited to compute the precise value of $\alpha$.)

@AsafKaragila already answered the question, but this is answering the follow-up question od @Reflecting_Ordinal in the comments to Asaf's question.

(i) $\alpha=\kappa_0+\beta$,

(i') there is a coded version of some such $M$ in $L_{\kappa_0+1}$; that is, there is a witnessing $M$ such that if $t$ is the $\Sigma_1$ theory of $M$ in parameters in $\beta\cup\{z\}$ where $z=(0^\#)^M$, and $t'$ is the theory which results by replacing $z$ with some constant symbol in $V_\omega\backslash\omega$, then $t'\in L_{\kappa_0+1}$,

(iii) $L_{\kappa_0}$ projects to $\omega$; i.e. $L_{\kappa_0+1}\cap\mathcal{P}(\omega)\not\subseteq L_{\kappa_0}$.

Note now that it follows that if we minimize on $\alpha$ first, giving say $\alpha'$, with some such $M\in L_{\alpha'+1}$, and then take the least witness $\beta'$ for $\mathrm{OR}^M$, then $\alpha'=\alpha$ and $\beta'=\beta$. For otherwise $\alpha'<\alpha$ but $\beta<\beta'$. But certainly $L_{\beta'}\models\mathrm{ZF}$, but the first ZF level beyond $L_\beta$ is well past $L_{\kappa_0+\beta}$, so $\alpha<\beta'\leq\alpha'$, contradiction.

So we have $\gamma=\kappa_0$. This gives $\alpha\leq\kappa_1+\beta$, where $\kappa_1$ is the least admissible $>\kappa_0$. But this doesn't suffice for (i).

Remark: Note that it also follows that $\kappa_0\notin\mathrm{wfp}(D)$ (otherwise we get surjection of $\omega$ onto $\beta$ definable over $M$), so $\kappa_0=\mathrm{wfp}(D)$.

Now for (i'): Because $L_{\kappa_0}$ projects to $\omega$, there is a surjection $\pi:\omega\to L_{\kappa_0}$ which is definable over $L_{\kappa_0}$. Therefore, we can fix a surjection $\sigma:\omega\to \mathscr{D}$, where $\mathscr{D}=$ the set of all dense subsets of $\mathbb{P}=\mathrm{Coll}(\omega,\beta)$ which are boldface-$\Sigma_2^{L_{\kappa_0}}$-definable, and such that $\sigma$ is definable over $L_{\kappa_0}$. Let $G$ be the filter $\subseteq\mathbb{P}$ which results by meeting all the sets in $\mathscr{D}$, one by one in the usual way, using the surjection $\sigma$. So $G\subseteq L_\beta$ and $G$ is definable over $L_{\kappa_0}$, so $G\in L_{\kappa_0+1}$. Now because $G$ meets enough dense sets, $L_{\kappa_0}[G]$ is also admissible, and contains a real $x$ coding $\beta$ (just given by $G$), such that letting $\tau:\omega\to\beta$ be the corresponding surjection, then $\tau$ is in $L_{\kappa_0}[G]$. Therefore like before, there is a model $M$ of the desired form coded by a real $m$, such that $m$ is definable from parameters over $L_{\kappa_0}[G]$, and such that the coding of ordinals ${<\beta}$ given by $m$ agrees with $\tau$; in fact, we can take $m$ to be $\Sigma_1\wedge\Pi_1$-definable from parameters over $L_{\kappa_0}[G]$, considering the complexity of the left-most-branch which yields $m$. But using $G$ and the $(\Sigma_1,\Pi_1)$-forcing relation over $L_{\kappa_0}$ (for $\mathbb{P}$), and names for the relevant parameters, we can define $m$ over $L_{\kappa_0}$, and also the theory $t'$ as described in (i'), associated to the model $M$. This gives (i').

Finally for (i), we get $\alpha\leq\kappa_0+\beta$ by (i'), as it takes at most $\beta$ steps to transitivize $m$. But actually, it takes exactly $\beta$ steps, because $(0^\#)^M$ is a real which is in $M$ which first appers in $L_{\kappa_0+1}$, and so note that the sets of the form $\ldots\{\{(0^\#)^M\}\}\ldots$ (with ${<\beta}$-many nested pairs of brackets, wellfounded), which are all in $M$, take $\beta$ stages of construction after $L_{\kappa_0+1}$ to produce. Therefore $\alpha=\kappa_0+\beta$. (But really, the more important ordinal here is $\kappa_0$.)

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Farmer S
Farmer S
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@AsafKaragila already answered the question, but this is addressing the follow-up question od @Reflecting_Ordinal in the comments to Asaf's question.

Let $\beta$ be least such that there is a transitive $M$ modelling "$V=L[0^\#]$" and $\alpha$ least such that there is such an $M$ in $L_{\alpha+1}$. (In the question, $\alpha$ was minimized first, whereas I have minimized $\beta$ first. In the end it actually gives the same $\alpha,\beta$.) Let $\kappa_0$ be least such that $\beta<\kappa_0$ and $L_{\kappa_0}$ is admissible, and $\kappa_1$ least such that $\kappa_0<\kappa_1$ and $L_{\kappa_1}$ is admissible. Then:

(i) $\kappa_0\leq\alpha\leq\kappa_1+\beta$,

(ii) $L_{\kappa_0}\models$"$\beta$ is inaccessible", so $\mathcal{H}_\beta^{L_{\kappa_0}}=L_{\beta}$,

(iii) $L_{\kappa_0}$ projects to $\omega$; i.e. $L_{\kappa_0+1}\cap\mathcal{P}(\omega)\not\subseteq L_{\kappa_0}$.

(I'm not sure where exactly $\alpha$ sits in the interval $[\kappa_0,\kappa_1+\beta]$.)

Note now that it follows that if we minimize on $\alpha$ first, giving say $\alpha'$, with some such $M\in L_{\alpha'+1}$, and then take the least witness $\beta'$ for $\mathrm{OR}^M$, then $\alpha'=\alpha$ and $\beta'=\beta$. For otherwise $\alpha'<\alpha$ but $\beta<\beta'$. But certainly $L_{\beta'}\models\mathrm{ZF}$, but the first ZF level beyond $L_\beta$ is well past $L_{\kappa_1+\beta}$, so $\alpha<\beta'\leq\alpha'$, contradiction.

Proof: Let $\gamma$ be least such that $\beta\leq\gamma$ and $L_\gamma$ projects to $\omega$. Let $\delta$ be least such that $L_\delta$ is admissible and $\delta>\gamma$. Because $\beta$ is countable in $L_\gamma$, it is a $\Sigma^1_1(x)$ statement about some real $x\in L_\gamma$ that there is a model $M$ modelling "$V=L[0^\#]$" whose ordinals are isomorphic to $\beta$. Because $L_\gamma$ is admissible, we therefore get a real $y$ coding such a model in $L_{\gamma+1}$. Then it takes at most $\beta$ further steps of construction to convert $y$ into a transitive version $M$, giving $M\in L_{\gamma+\beta+1}$. So $\alpha\leq\gamma+\beta$.

Now let's verify (ii). For this, working inside one of the models $M$ of interest, we can iterate $0^\#$ out (considered as a mouse with external measure) through $\beta$, and can compute the direct limit $D$ of that iteration. Note that $\beta+1$ is in the wellfounded part $\mathrm{wfp}(D)$ of $D$ (as $\beta$ is just the thread of the critical points), and therefore $L_{\kappa_0}\subseteq\mathrm{wfp}(D)$. Since this is all definable over $M$, and $M\models\mathrm{ZF}$, it follows that $L_{\kappa_0}$ contains no bounded subsets of $\beta$ that are not already in $M$, i.e. we have $\mathcal{P}({<\beta})\cap L_{\kappa_0}\subseteq M$, and it also follows that there is no singularization of $\beta$ in $L_{\kappa_0}$. But if $L_{\kappa_0}$ contained any subsets of some $\tau<\beta$ that are missing from $L_\beta$, then in fact it contains a surjection from $\tau\to\beta$, which gives a subset of $\tau$ missing from $M$, contradiction. So $\beta$ is inaccessible in $L_{\kappa_0}$. This gives (ii).

Now for (iii): Suppose that $L_{\kappa_0}$ does not project to $\omega$. Then working in $L_{\kappa_0+1}$, taking $n<\omega$ as large as we like, we can form a countable $\Sigma_n$-elementary substructure $N$ of $L_{\kappa_0}$, and so ther is some $\kappa_0'<\omega_1^{L_{\kappa_0}}=\omega_1^{L_{\beta}}<\beta$ and $\Sigma_n$-elementary $\pi:L_{\kappa_0'}\to L_{\kappa_0}$. But $L_{\kappa_0}\models$"It is forced by $\mathrm{Coll}(\omega,\beta)$ that in the generic extension, a real $y$ can be defined from parameters, such that $y$ codes a model $P$ satisfying $V=L[0^\#]$, whose ordinals are isomorphic to $\beta$"; this is just like in the first paragraph of the proof, except now we use a generic enumeration of $\beta$ instead of the one we got in $L_{\gamma+1}$. We can take $n$ large enough that this statement reflects to $L_{\kappa_0'}$. But then we can take a generic $G$ over $L_{\kappa_0'}$, and find a witnessing real $y\in L_{\kappa_0'}[G]$, and hence there is a model with $\beta'<\beta$, contradicting minimality. This gives (iii).

So we have $\gamma=\kappa_0$, completing the proof.

Remark: Note that it also follows that $\kappa_0\notin\mathrm{wfp}(D)$ (otherwise we get surjection of $\omega$ onto $\beta$ definable over $M$), so $\kappa_0=\mathrm{wfp}(D)$.