 Solution

TakeStart 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 $1\le j\le m-2$. 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 $S_3$.