= Solution
Apply part (a). The subgroup $H$ is finite because it lies in the finite set $S^{\lfloor n/2\rfloor}$. Conjugation gives a homomorphism
$$
C\longrightarrow\operatorname{Aut}(H).
$$
Its kernel $C_0$ has finite index in $C$ and centralizes $H$. Since $C_0H/H$ is a subgroup of the $O_d(1)$-step nilpotent group $C/H$, it is itself nilpotent of class $O_d(1)$. Hence
$$
\gamma_{s+1}(C_0)\subseteq H
$$
for some $s=O_d(1)$. As $C_0$ centralizes $H$, one more <group commutator> vanishes, so $\gamma_{s+2}(C_0)=\{1\}$. Thus $C_0$ is nilpotent of class at most $s+1=O_d(1)$.
Both $[G:C]$ and $[C:C_0]$ are finite, so
$$
\boxed{G\text{ has an }O_d(1)\text{-step nilpotent subgroup of finite index}.}
$$
This is the <Gromov theorem on groups of polynomial growth> in the form needed here.
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