 Solution

Every ordinal belowhas an ultrapower representative $f:\kappa \to {\kappa }^{+}$: by Łoś theorem, a representative below the successor of $j(\kappa )$ may be chosen below ${\kappa }^{+}$ on a set in the ultrafilter. Hence, in $V_{\lambda }$,The ultrapower is closed under $\kappa$-sequences, so it contains every subset of $\kappa$ and computes $2^{\kappa }$ correctly. By elementarity $M$ regards $j(\kappa )$ as measurable, hence strongly inaccessible by Question 1(b). ConsequentlyThus $V_{\lambda }$ has a set of cardinality at most $2^{\kappa }$ whose order type is $\eta >2^{\kappa }$, so