Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, it suffices to exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially, and there is a standard trick to this: take inertia at $31$. Let me call it $I$. This must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact $\mathbb{Q}$ has no extensions that are unramfied at all finite places — and it is intersects $N$ trivially, since $H_2/K$ is everywhere unramified, while the extension cut out by $I$ is totally ramified at $31$.

This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this from our group theoretic considerations!) and will, after a few minutes, spit out a slightly horrendous looking polynomial of degree $9$ over $L$ whose root generates the Hilbert class field of $L$: $$ x^9 + \tfrac{1}{256}(351\alpha^4 + 22842\alpha^2 + 999)x^7 + \tfrac{1}{256}(-9585\alpha^4 + 33210\alpha^2 + 567)x^6 + \tfrac{1}{256}(56133\alpha^4 + 756702\alpha^2 + 26973)x^5 + \tfrac{1}{128}(-14096673\alpha^4 - 289073286\alpha^2 - 9985113)x^4 + \tfrac{1}{256}(837980397\alpha^4 + 2627921070\alpha^2 + 89938917)x^3 + \tfrac{1}{64}(-525358953\alpha^4 + 150497910738\alpha^2 + 5208850071)x^2 + \tfrac{1}{128}(500734949193\alpha^4 - 5434147475802\alpha^2 - 188657086959)x + \tfrac{1}{128}(329428602877167\alpha^4 + 3754393943660730\alpha^2 + 129532294910295), $$ where $\alpha\in L$ has minimal polynomial $$x^6 + 93x^4 + 899x^2 + 31$$ over $\mathbb{Q}$. Its compositum with $K$, obtained by adjoining to $K$ a root of the same polynomial, must then be the Hilbert class field of $K$.

Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, that argument does not work, so instead we will exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially. There is a standard trick to this: take inertia at $31$. Let me call it $I$. It must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact that $\mathbb{Q}$ has no extensions that are unramified at all finite places — and it intersects $N$ trivially, since $H_2/K$ is everywhere unramified, while the extension cut out by $I$ is totally ramified at $31$.

This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this from our group theoretic considerations!) and will, after a few minutes, spit out a slightly horrendous looking polynomial of degree $9$ over $L$ whose root generates the Hilbert class field of $L$: $$ x^9 + \tfrac{1}{256}(351\alpha^4 + 22842\alpha^2 + 999)x^7 + \tfrac{1}{256}(-9585\alpha^4 + 33210\alpha^2 + 567)x^6 + \tfrac{1}{256}(56133\alpha^4 + 756702\alpha^2 + 26973)x^5 + \tfrac{1}{128}(-14096673\alpha^4 - 289073286\alpha^2 - 9985113)x^4 + \tfrac{1}{256}(837980397\alpha^4 + 2627921070\alpha^2 + 89938917)x^3 + \tfrac{1}{64}(-525358953\alpha^4 + 150497910738\alpha^2 + 5208850071)x^2 + \tfrac{1}{128}(500734949193\alpha^4 - 5434147475802\alpha^2 - 188657086959)x + \tfrac{1}{128}(329428602877167\alpha^4 + 3754393943660730\alpha^2 + 129532294910295), $$ where $\alpha\in L$ has minimal polynomial $$x^6 + 93x^4 + 899x^2 + 31$$ over $\mathbb{Q}$. Its compositum with $K$, obtained by adjoining to $K$ a root of the same polynomial, must then be the Hilbert class field of $K$.

Edit: Franz Lemmermeyer has found a much nicer polynomial that generates the same field over $L$, and therefore the same field over $K$: $$ x^9 - x^7 - 2x^6 + 3x^5 + x^4 + 2x^3 - x^2 + x - 3. $$

Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, it suffices to exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially, and there is a standard trick to this: take inertia at $31$. Let me call it $I$. This must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact $\mathbb{Q}$ has no extensions that are unramfied at all finite places — and it is disjoint from $N$ since $H_2/K$ is everywhere unramified.

This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this!) and will, after a few minutes, spit out a (slightly horrendous looking) polynomial of degree $9$ whose root generates the Hilbert class field of $L$. Its compositum with $K$ must then be the Hilbert class field of $K$.

Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, it suffices to exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially, and there is a standard trick to this: take inertia at $31$. Let me call it $I$. This must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact $\mathbb{Q}$ has no extensions that are unramfied at all finite places — and it is intersects $N$ trivially, since $H_2/K$ is everywhere unramified, while the extension cut out by $I$ is totally ramified at $31$.

This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this from our group theoretic considerations!) and will, after a few minutes, spit out a slightly horrendous looking polynomial of degree $9$ over $L$ whose root generates the Hilbert class field of $L$: $$ x^9 + \tfrac{1}{256}(351\alpha^4 + 22842\alpha^2 + 999)x^7 + \tfrac{1}{256}(-9585\alpha^4 + 33210\alpha^2 + 567)x^6 + \tfrac{1}{256}(56133\alpha^4 + 756702\alpha^2 + 26973)x^5 + \tfrac{1}{128}(-14096673\alpha^4 - 289073286\alpha^2 - 9985113)x^4 + \tfrac{1}{256}(837980397\alpha^4 + 2627921070\alpha^2 + 89938917)x^3 + \tfrac{1}{64}(-525358953\alpha^4 + 150497910738\alpha^2 + 5208850071)x^2 + \tfrac{1}{128}(500734949193\alpha^4 - 5434147475802\alpha^2 - 188657086959)x + \tfrac{1}{128}(329428602877167\alpha^4 + 3754393943660730\alpha^2 + 129532294910295), $$ where $\alpha\in L$ has minimal polynomial $$x^6 + 93x^4 + 899x^2 + 31$$ over $\mathbb{Q}$. Its compositum with $K$, obtained by adjoining to $K$ a root of the same polynomial, must then be the Hilbert class field of $K$.