= Solution
The <cumulative hierarchy> is defined by <transfinite recursion>:
$$
V_0=\varnothing,\qquad V_{\alpha+1}=\mathcal P(V_\alpha),\qquad
V_\lambda=\bigcup_{\beta<\lambda}V_\beta\quad(\lambda\text{ a limit ordinal}).
$$
The <rank of a set> is $\operatorname{rk}(x)=\sup_{y\in x}(\operatorname{rk}(y)+1)$, equivalently the least $\alpha$ such that $x\subseteq V_\alpha$. The standard hierarchy properties give $\operatorname{rk}(V_\alpha)=\alpha$, so \b[every <ordinal> occurs as a set rank]; the <ordinal> $\alpha$ itself is another example of rank $\alpha$.
Let $x$ be transitive of rank $\alpha$, and fix $\beta<\alpha$. The supremum formula gives an element of $x$ whose rank is at least $\beta$. Choose such an element $y$ of least possible rank. If $\operatorname{rk}(y)>\beta$, the same supremum formula gives $z\in y$ with rank at least $\beta$. Transitivity gives $z\in x$, and membership strictly decreases rank, contradicting the choice of $y$. Thus \b[$x$ contains an element of every rank below $\alpha$].
A nonempty finite set has rank the maximum of finitely many <successor ordinals>, hence a <successor ordinal>; the empty set has rank zero. Conversely, for any $\beta$, the singleton $\{V_\beta\}$ has rank $\beta+1$. Thus
$$
\boxed{\text{finite-set ranks are exactly }0\text{ and all successor ordinals}.}
$$
A finite <transitive set> of rank $\alpha$ contains, by the proved property, distinct elements of every rank $\beta<\alpha$. Finiteness therefore forces $\alpha<\omega$. Every finite <ordinal> is itself a finite <transitive set> of that rank. Consequently
$$
\boxed{\text{finite-transitive-set ranks are exactly the finite ordinals}.}
$$
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