= Solution
A suitable <relative condensation lemma> fixes the base pointwise. Let $\alpha$ be a nonzero <limit ordinal>, and let $Y\prec(L_\alpha(X),\in)$ be an <elementary substructure> with
$$
X\cup\{X\}\subseteq Y.
$$
Then the <Mostowski collapse theorem> gives an isomorphism $\pi:Y\to T$ onto a <transitive set>, and
$$
\boxed{T=L_\beta(X)\quad\text{for a nonzero limit ordinal }\beta\leq\alpha,\qquad\pi\!\upharpoonright X=\operatorname{id}_X.}
$$
For the collapse to apply, the restricted membership relation is well-founded because it is actual membership. It is extensional: if two members of $Y$ differ, an element distinguishing them in $L_\alpha(X)$ can be chosen in $Y$ by <elementary substructure> structure. Since $X$ is transitive and all its members are in $Y$, <transfinite induction> on their <rank of a set> shows that $\pi$ fixes every member of $X$, and then $\pi(X)=X$.
By the preceding <relative constructible level recognition>, $L_\alpha(X)\models\Phi(X)$. <Elementary substructure> agreement transfers this to $Y$, and the collapse isomorphism transfers it to $T$. Thus $T\models\Phi(X)$ and $T=L_\beta(X)$ for a nonzero <limit ordinal> $\beta$.
For the bound, let $\rho=X\cap\operatorname{Ord}$. The <relative constructible hierarchy> has ordinal height $L_\gamma(X)\cap\operatorname{Ord}=\rho+\gamma$: at a successor the <definable power set> adds exactly the previous ordinal height as a new <ordinal>, and at limits take unions. The order type of $Y\cap\operatorname{Ord}$ is at most $\rho+\alpha$, so $\rho+\beta\leq\rho+\alpha$; strict increase of <ordinal addition> in the right argument gives $\beta\leq\alpha$. The requirement $X\subseteq Y$ is essential to this version: merely having $X\in Y$ need not make the collapse fix the base.
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