= Solution
Use the standard convention that a <strongly inaccessible cardinal> is uncountable, regular and strong limit. \b[The printed explanatory definition omits uncountability]: literally it also admits $\kappa=\omega$, for which no finite cardinal is a <worldly cardinal>. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended <strongly inaccessible cardinal> convention $\kappa>\omega$.
First $|V_\gamma|<\kappa$ for every $\gamma<\kappa$. At successor stages this follows from the <strong limit cardinal> property, and at limits from the <regular cardinal> property. Consequently $V_\kappa\models\mathsf{ZFC}$. In particular, for <Axiom schema of replacement>, a domain $a\in V_\kappa$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the <rank of a set> over its elements is below $\kappa$ by <regular cardinal> structure. All other axioms, including <axiom of choice>, have their witnesses at bounded ranks below $\kappa$; uncountability supplies <axiom of infinity>.
Enumerate the <first-order formulas> as $\varphi_0,\varphi_1,\ldots$. For each finite collection, reflection inside the set structure $V_\kappa$ gives a <closed unbounded subset> of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_\gamma$, and the supremum of their least witness ranks remains below $\kappa$ by <regular cardinal> structure. Closing under these bounds and taking increasing countable limits gives the usual <reflection theorem for definable hierarchies> argument. The countable intersection of these <closed unbounded subsets> is still <closed unbounded> because $\kappa$ is regular and uncountable. At each resulting $\alpha$,
$$
(V_\alpha,\in)\prec(V_\kappa,\in),\qquad V_\alpha\models\mathsf{ZFC}.
$$
The infinite <cardinal numbers> below $\kappa$ also form a <closed unbounded subset>: they are unbounded because $\lambda^+\leq2^\lambda<\kappa$ for infinite $\lambda<\kappa$, and a supremum of increasing <cardinal numbers> is a <cardinal number>. Intersect the two <closed unbounded subsets>. Every member of the intersection is a <worldly cardinal>, and an unbounded <subset> of a <regular cardinal> $\kappa$ has size $\kappa$. Hence
$$
\boxed{\left|\{\alpha<\kappa:\alpha\text{ is a cardinal and }V_\alpha\models\mathsf{ZFC}\}\right|=\kappa.}
$$
In fact this proves the stronger <worldly cardinals below an inaccessible cardinal> result that these <worldly cardinals> contain a <closed unbounded subset> of $\kappa$.
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