= Solution
Every ordinal below
$$
\eta=(j(\kappa)^+)^M
$$
has 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 $V_\lambda$,
$$
|\eta|\leq(\kappa^+)^\kappa=2^\kappa.
$$
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). Consequently
$$
2^\kappa<j(\kappa)<\eta.
$$
Thus $V_\lambda$ has a set of cardinality at most $2^\kappa$ whose order type is $\eta>2^\kappa$, so
$$
V_\lambda\models\text{“$\eta$ is not a cardinal”.}
$$
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