= Solution
The class $V$ is transitive because all its levels are transitive. Let $a\in V$ be nonempty. There is a least <ordinal> $\alpha$ for which $a\cap V_\alpha\ne\varnothing$: existence follows by choosing an element of $a$, which lies in $V$, and then minimizing within some witnessing <ordinal> bound. The least <ordinal> is neither zero nor a limit, so write $\alpha=\beta+1$. Choose $x\in a\cap V_{\beta+1}$. Then $x\subseteq V_\beta$, while minimality gives $a\cap V_\beta=\varnothing$. Thus $a\cap x=\varnothing$. Also $x\in V$, proving \b[<Foundation> inside $\langle V,\in\rangle$].
By the given hypothesis this same class satisfies every axiom of $ZF^*$, and hence it satisfies every axiom of $ZF$. Translate any finite formal $ZF$ proof by restricting each <quantifier> to the definable class $V$. Within $ZF^*$ the translated axioms have proofs, logical steps remain valid, and the translated contradiction is still a contradiction. Substituting those axiom proofs therefore converts the original derivation effectively into a finite derivation of contradiction from $ZF^*$. Equivalently,
$$
\boxed{\operatorname{Con}(ZF^*)\Longrightarrow\operatorname{Con}(ZF).}
$$
This is the syntactic relative-consistency construction; it does not assume that $V$ is a set or that it equals the whole ambient universe.
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