= Solution
Yes. Start with the model $M_0$ constructed in part a. If $M_0$ has no internally <strongly inaccessible cardinal> above $\kappa$, put $M=M_0$. Otherwise let $\delta$ be the least ordinal above $\kappa$ that $M_0$ regards as strongly inaccessible, and put
$$
M=(V_\delta)^{M_0}.
$$
In the second case $M\models\mathrm{ZFC}$ because $M_0$ regards $\delta$ as inaccessible. The measure witnessing that $\kappa$ is <measurable cardinal>[measurable] has rank below $\kappa+3<\delta$, so it still belongs to $M$. In both cases $M$ is a <transitive set> of cardinality $\kappa$, contains $V_\kappa$, and has no internally inaccessible ordinal strictly between $\kappa$ and its height.
We verify <set-theoretic absoluteness>[absoluteness] for every ordinal $\alpha\in M$. If $\alpha<\kappa$, then $M$ and $V_\lambda$ both contain $V_{\alpha+1}$ and therefore compute all subsets and functions relevant to strong inaccessibility in the same way. At $\alpha=\kappa$, both models see a <measurable cardinal> and hence an inaccessible cardinal. Finally, if $\kappa<\alpha\in M$, then $M$ says that $\alpha$ is not inaccessible by construction. The larger model $V_\lambda$ cannot say that it is inaccessible, because strong inaccessibility is downward absolute to a transitive model of ZFC: any failure visible in the smaller model remains a failure in the larger one, while ambient inaccessibility would force internal inaccessibility. Hence “$\alpha$ is inaccessible” is absolute between $M$ and $V_\lambda$.
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