= Solution
The <cumulative hierarchy> is defined by <transfinite recursion>:
$$
\boxed{V_0=\varnothing,\qquad V_{\alpha+1}=\mathcal P(V_\alpha),\qquad V_\lambda=\bigcup_{\alpha<\lambda}V_\alpha\ \text{for limit }\lambda.}
$$
Prove transitivity and nesting together by <transfinite induction>. The empty set is transitive. If $V_\alpha$ is transitive, each $x\in V_\alpha$ satisfies $x\subseteq V_\alpha$, so $V_\alpha\subseteq\mathcal P(V_\alpha)$. If $y\in x\in\mathcal P(V_\alpha)$, then $y\in V_\alpha\subseteq\mathcal P(V_\alpha)$; thus the successor stage is transitive. At a limit, if $y\in x\in V_\lambda$, choose an earlier stage containing $x$ and apply its transitivity. The union is transitive and contains every earlier stage. These observations also give $V_\alpha\subseteq V_\beta$ for all $\alpha\le\beta$.
By <Axiom of foundation> and well-founded recursion, define the <rank of a set> by $\operatorname{rank}(x)=\sup\{\operatorname{rank}(y)+1:y\in x\}$. <Axiom schema of replacement> ensures this is an <ordinal>. Induction shows every element of $x$ lies in $V_{\operatorname{rank}(x)}$, so
$$
\boxed{x\in V_{\operatorname{rank}(x)+1}.}
$$
Thus every set appears in the hierarchy; the argument uses foundation, not the existence of a set containing all sets.
If $\operatorname{rank}(x)=\beta$, every subset of $x$ has rank at most $\beta$, while $x$ itself belongs to $\mathcal P(x)$. Consequently
$$
\boxed{\operatorname{rank}(\mathcal P(x))=\beta+1.}
$$
Only successor <ordinals> occur. Conversely $\operatorname{rank}(V_\beta)=\beta$, so choosing $x=V_\beta$ realizes every successor $\beta+1$. \b[The exact answer is all successor <ordinals>, excluding zero and nonzero limit <ordinals>.]
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