An electron falls from rest through a vertical distance h in a...

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

Initially mass is at rest, so $u = 0$.

$h = 0 + \dfrac{1}{2}\left(\dfrac{qE}{m}\right){t}^{2} ⇒ t = \sqrt{\dfrac{2hm}{qE}} ⇒ t ∝ \sqrt{m}$

as mass of proton is heavier than electron, hence electron will take less time.