An electron falls from rest through a vertical distance h in a uniform and vertically upward directed…

 Acceleration of charge $q$ of mass $m$ in uniform electric field is given as $a=\frac{qE}{m}$.
Applying second equation of motion, $S=ut+\frac{1}{2}a{t}^2$.

Initially mass is at rest, so $u=0$.
$h=0+\frac{1}{2} \left(\frac{qE}{m}\right)t^2\Rightarrow t=\sqrt{\frac{2hm}{qE}}\Rightarrow t\propto \sqrt{m}$
as mass of proton is heavier than electron, hence electron will take less time.