Let $G$ be a free group generated by $\{a_1,a_2,...,a_n\}$,$[G,G]$ denote the subgroup of $G$ generated by commutators $\{aba^{-1}b^{-1}\mid a,b\in G\}$
hence we know $G/[G,G]$ is a free abelian group, generated by $\overline{a_1},...,\overline{a_n}$ (where $\overline{a}_i $ is image under the canonical projection $\pi:G\to G/[G,G]$ ).
assume normal subgroup $N\subset G$ is the least normal subgroup containing the element $\{a_1^2a^2_2...a_n^2\}$.where $[G,G]N$ denote the subgroup $\{an\mid a\in [G,G],n\in N\}\subset G$
Now we want to study what $\frac{G/[G,G]}{[G,G]N/[G,G]}$ is,here gives the result ,letting $G' = $ free abelian group generated by $\overline{a_1},...,\overline{a_n}$
hence we have :
$$\frac{G/[G,G]}{[G,G]N/[G,G]} = G'/\left<\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2\right> \tag{*}$$
I have no idea how to get (*),the point is why $\left<\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2\right> = [G,G]N/[G,G]$
I can do as follows since $ \pi(a_1^2a_2^2...a_n^2)\subset \pi(N) =[G,G]N/[G,G] $
We have $\left<\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2\right> \subset[G,G]N/[G,G]$
Second $N$ is least normal subgroup of $\{a_1^2a^2_2...a_n^2\}$ by surjective of $\pi$ we have $\pi(N)$ is the least normal subgroup of $\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2$ since if exist another normal subgroup $K\subset G/[G,G]$ containing $\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2$ we have $\pi(N) \subset \pi(\pi^{-1}(K))\subset K$ which means $\pi(N)$ is the least normal one,but $\left<\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2\right>$ itself is normal we have $\pi(N) = \left<\overline{a_1}^2\overline{a_2}^2...\overline{a_n}^2\right>$