= Solution
Put $\Phi=\Phi(G)$ and let $H$ be the full preimage of $F(G/\Phi)$ in $G$. The preceding part gives $\Phi\le F(G)$, so the normal nilpotent <image> $F(G)/\Phi$ lies in $F(G/\Phi)$.
For the reverse inclusion, choose a <Sylow subgroup> $P$ of $H$. Its <image> $P\Phi/\Phi$ is the unique <Sylow subgroup> of the <nilpotent group> $H/\Phi$, hence a <characteristic subgroup> there and normal in $G/\Phi$. Thus $P\Phi\trianglelefteq G$. The <subgroup> $P$ is also a <Sylow subgroup> of $P\Phi$: the p-part of $\Phi$ lies in $P$, because $P\cap\Phi$ is a <Sylow subgroup> of the <normal subgroup> $\Phi$ of $H$. Apply the <Frattini argument> to $P\Phi$:
$$
G=(P\Phi)N_G(P)=\Phi N_G(P).
$$
The <Frattini subgroup> property forces $N_G(P)=G$. Hence every <Sylow subgroup> of $H$ is normal, so $H$ is a <nilpotent group> and $H\le F(G)$. Together the inclusions prove
$$
\boxed{F(G/\Phi(G))=F(G)/\Phi(G).}
$$
This is <Frattini lifting of nilpotence>; simply calling an extension of <nilpotent groups> nilpotent would not establish it.
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