 Solution

Let $U$ witness that $\kappa$ is measurable. Regularity is given, so it remains to prove the strong limit cardinal property. First, every $A\in U$ has cardinality $\kappa$: if $|A|<\kappa$, thenby nonprincipality and $\kappa$-completeness, contradicting $A\in U$.

Suppose $\lambda <\kappa$ and $2^{\lambda }\ge \kappa$. Choose an injection $f:\kappa \to \mathcal{P}(\lambda )$. For each $\xi <\lambda$, exactly one oflies in $U$. Their chosen intersection lies in $U$ by $\kappa$-completeness. On that intersection every $f(\alpha )$ is the same subset of $\lambda$, contradicting injectivity because every member of $U$ has size $\kappa$. Thus $2^{\lambda }<\kappa$, and $\kappa$ is strongly inaccessible.