= Solution
Take
$$
\boxed{G=S_3\times C_2^{m-2}\quad(m\geq2).}
$$
Start with $1<A_3<S_3$, whose factors are $C_3$ and $C_2$, and then append the $m-2$ coordinate factors of the elementary <Abelian group> $C_2^{m-2}$. More explicitly, append $S_3\times C_2^j$ for $1\leq j\leq m-2$. Each subgroup is normal in the next, and the factors consist of one $C_3$ and $m-1$ copies of $C_2$. Thus there are exactly $m$ finite cyclic factors. The <group> is non-<Abelian> because it contains $S_3$.
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