Let $G$ be a finite solvable group and $N$ be a normal subgroup of it which is an elementary abelian 2-group. Suppose that $G/N\cong \mathbb{Z}_2\times \mathbb{Z}_2$ and $|C_G(x)|=16$ for any $x\in G-N$. What can we say about $G$? Is it abelian? Is it elementary?
We know that since $G$ is a 2-group, it is nilpotent.
I would provide some examples, on which you may build some expectations:
$G=C_2\times C_2\times D_4$, where $D_4$ is the dihedral group of $8$ elements. $N=C_2\times C_2\times C_2$ is the center of $G$. Since $D_4$ mod its center is $C_2\times C_2$, the factor group condition holds. For $x \in G-N$, $|C_G(x)|$ contains the $N$ and $x$, so its size is strictly larger than $8$. Also, $|C_G(x)|\neq 32$, otherwise $x$ is in the center. So $|C_G(x)|=16$ for any element not in the center.
$G$ can be replaced by $C_2\times C_2\times Q_8$ ($Q_8$ is the quarernion group). The reasons are the same.
There are three more groups which can play the role of $G$, and $N$ their (respective) centralizers. The structure of $N$ and $G/N$ can be read off from the webpage; and for the size of the centralizer, you can use the formula $|C(x)|=\sum_{χ}|χ(x)|^2$, where $x$ is a conjugacy class and $χ$ runs over the group characters.
