= Solution
For a set $A$, its <definable power set> is
$$
\operatorname{Def}(A)=\{\{x\in A:\langle A,\in\rangle\models\varphi(x,p_1,\ldots,p_n)\}:\varphi\text{ a formula},\ p_1,\ldots,p_n\in A\}.
$$
Only finitely many parameters occur in each definition. Satisfaction for the set structure is definable, so this specifies a uniformly definable class function assigning a set to every set $A$, with $\operatorname{Def}(A)\subseteq\mathcal P(A)$. Every subset definable over the structure with its parameters occurs. The full set $A$ itself occurs by the formula $x=x$. If $A$ is transitive, then $A\subseteq\operatorname{Def}(A)$, since any $p\in A$ is defined as a subset of $A$ by $x\in p$. In that case $\operatorname{Def}(A)$ is transitive as well: a member of one of its elements lies in $A$, hence in $\operatorname{Def}(A)$.
The <constructible hierarchy> and universe are
$$
\boxed{L_0=\varnothing,\quad L_{\alpha+1}=\operatorname{Def}(L_\alpha),\quad L_\lambda=\bigcup_{\beta<\lambda}L_\beta,\quad L=\bigcup_{\alpha\in\operatorname{Ord}}L_\alpha.}
$$
The displayed properties prove by induction that levels are transitive and increasing, so $L$ is transitive. It contains every <ordinal>: inductively $L_\alpha\cap\operatorname{Ord}=\alpha$, and the subset of $L_\alpha$ consisting of <ordinals> is definable there, giving $\alpha\in L_{\alpha+1}$. Ordinalhood is absolute here because the levels are transitive. Both the level function and membership in $L$ are definable in $ZF$. This is Gödel's <constructible universe>, which can contain fewer subsets than the full universe.
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