= Solution
Fix $\varepsilon>0$ and set $d=2/\varepsilon$. Suppose, towards a contradiction, that
$$
D=\operatorname{diam}_S(G)>\max\{|G|^\varepsilon,\lambda\},
$$
where $\lambda$ will absorb constants depending only on $\varepsilon$. Put $n=\lfloor|G|^\varepsilon\rfloor$. Once $\lambda$ is large enough, $n\geq N(d)$, $n<D$, and
$$
|S^n|\leq|G|\leq n^d|S|.
$$
Part (a) yields $H\trianglelefteq C<G$ with $[G:C]=O_\varepsilon(1)$, $H\subseteq S^{\lfloor n/2\rfloor}$, and $C/H$ nilpotent of class $O_\varepsilon(1)$.
The <subgroup core> $\bigcap_{g\in G}gCg^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a <simple group>, the core is either $\{1\}$ or $G$. In the first case $|G|\leq [G:C]!=O_\varepsilon(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.
Now $H\trianglelefteq G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor}$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a <nilpotent group>. A nontrivial finite nilpotent group has nontrivial <center of a group>; simplicity would force that center to be all of $G$, making $G$ <Abelian>. This contradicts the assumption that $G$ is non-abelian. Consequently
$$
\boxed{\operatorname{diam}_S(G)\leq\max\{|G|^\varepsilon,\lambda(\varepsilon)\}.}
$$
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