#### Answer

$v=1.48\times10^{15}m/s$

#### Work Step by Step

$E=1000 V/m$
Mass = $m= 9.1\times10^{-31}kg$
The electron accelerates in vertically downward direction.
$a_{e}=\frac{E}{mass}=\frac{1000}{9.1\times10^{-31}}=1.09\times10^{33}m/s^{2}$
Now from the equation,
$v^{2}=u^{2}+2aS$
with $u=0$, $a$ is calculated above, $S=0.1cm$, we can calculate $v$
So, $v=1.48\times10^{15}m/s$