$a=\frac{F}{m}=\frac{q E}{m}$

$a_{A}=\frac{1}{4} a_{B}$

Now $v^{2}=u^{2}+2 a x$

since, $x$ is the same for both the particles.

$\Rightarrow \quad v_{A}^{2}=0+2 a_{A} x$ and $v_{B}^{2}=0+2 a_{B} x$

$\Rightarrow \frac{v_{A}^{2}}{v_{B}^{2}}=\frac{a_{A}}{a_{B}}=\frac{1}{4} \Rightarrow \frac{v_{A}}{v_{B}}=\frac{1}{2}$