 Solution

An strongly inaccessible cardinal $\kappa$ has the Keisler extension property when there is a proper transitive set $X⊋V_{\kappa }$ such that

Suppose $\kappa$ is strongly inaccessible and has this property. Because $X$ properly extends the transitive set $V_{\kappa }$, it contains $\kappa$. Strong inaccessibility of $\kappa$ is downward absolute from the ambient universe to the transitive set $X$: any internal witness that $\kappa$ is countable, singular, or not a strong limit would also be an ambient witness. Hencewith $\kappa$ as a witness. Since $V_{\kappa }\prec X$, the same sentence holds in $V_{\kappa }$. Its witness is an ordinal $\mu <\kappa$. The set $V_{\kappa }$ contains $V_{\mu +1}$, so it computes all subsets of cardinals below $\mu$ correctly; strong inaccessibility of $\mu$ is therefore absolute between $V_{\kappa }$ and the universe. Thus there is a strongly inaccessible $\mu <\kappa$, and $\kappa$ cannot be the least strongly inaccessible cardinal.