The force experienced by the electron due to the electric field is given by:
[tex]F=qE=(1.6 \cdot 10^{-19}C)(800 N/C)=1.28 \cdot 10^{-16}N[/tex]
The work done by the field over a distance of d=4 cm=0.04 m is equal to the energy acquired by the electron during its motion:
[tex]W=\Delta E= Fd=(1.28 \cdot 10^{-16}N)(0.04 m)=5.12 \cdot 10^{-18}J[/tex]
Since the electron starts from rest, its initial kinetic energy is zero, so this energy acquired is equal to the final kinetic energy of the electron:
[tex]K=5.12 \cdot 10^{-18}J[/tex]
And from the formula
[tex]K= \frac{1}{2}mv^2 [/tex]
we can find the speed of the electron:
[tex]v= \sqrt{ \frac{2K}{m} } = \sqrt{ \frac{2(5.12 \cdot 10^{-18} J)}{9.11 \cdot 10^{-31} kg} }=3.35 \cdot 10^6 m/s [/tex]
