 Solution

False. The reasonable sets form the transitive set $H_{{\omega }_1}$, but this structure fails the Axiom of power set. The set $\omega$ is reasonable, and every subset of $\omega$ is itself countable and therefore reasonable. Its full power set $\mathcal{P}(\omega )$ is uncountable, so it is not reasonable. No member of $H_{{\omega }_1}$ can therefore contain internally every subset of $\omega$ that belongs to $H_{{\omega }_1}$. Thus the reasonable sets do not form a model of ZFC.