A uniform electric field of strength E exists in a region. An electron (charge -e, mass m) enters a points A perpendicular to x-axis with velocity V . It moves through the electric field & exits at point B. Then: 

 

As particle's velocity along Y axis ls not changing so it means electric field is not having any component in Y direction. As particle is having non-zero x-component of displacement in positive direction so it means electric field is along negative X-axis. From given information, $\mathrm{d}= \mathrm{v} \times \mathrm{t}$where t is the time taken by electron to move from A to B. 

So, $a=\frac{1}{2}\times \frac{eE}{m}{t}^2$

Solving above equation we get, $E=-\frac{2am{v}^2}{ed}\hat{i}$

Rate of work done is given by $\mathrm{P}=\overrightarrow{\mathrm{F}}\cdot \overrightarrow{\mathrm{V}}= F{V}_x=\left(eE\right)(\frac{eE}{m}.\frac{d}{v})$