As $qE=\frac{mv^{2}}{r}$,
$r=\frac{mv^{2}}{qE}$
$=\frac{2K}{qE}$
(where $K=\frac{1}{2}mv^{2}$
i.e, $r \propto\frac{1}{q}$
Since electron and proton have equal charge, r is the same for both of them
Here, $r$ depends on $K$ (which is given to be the same for both the particles) and as such not on their masses