Answer
$E=1.02\times10^{-2}\frac{N}{C}$
Work Step by Step
We know that
$F=q_eE$
but $F=ma$
so
$ma=q_eE$
or
$E=\frac{ma}{q_e}$
putting the relevant values of electron, we get
$E=\frac{(9.11\times10^{-31}kg)(1.8\times10^9\frac{m}{s^2})}{-1.6\times10^{-19}C}$
$E=1.02\times10^{-2}\frac{N}{C}$
