Given here, kinetic energy $\frac{1}{2}e=\frac{1}{2}m{u}_x^2\Rightarrow u_x=\sqrt{\frac{e}{m}}$

Acceleration of electron due to electric field is $a=\frac{eE}{m}$, where, $e$ is charge on electron.

Using equation of motion, $s=u_{x}t+\frac{1}{2}{a}_x{t}^2$  along $x$ direction, we get

$$\begin{gathered} L=u_{x}t+0\left(\because a_x=0\right) \\ \Rightarrow 10\times {10}^{-2}=u_{x}t \\ \Rightarrow t=\frac{{10}^{-1}}{u_x} ....\left(1\right) \end{gathered}$$

Using equation of motion, $v_y=u_y+a_{y}t$ along y direction, we get

$v_y=0+\frac{eE}{m}t=\frac{eE}{m}t ...\left(2\right)$

From equation $\left(1\right) \mathrm{and} \left(2\right)$, we have

$v_y=\frac{eE}{m}\left(\frac{{10}^{-1}}{u_x}\right)$

Now, $\tan \theta =\frac{v_y}{v_x}=\frac{eE}{m}\left(\frac{{10}^{-1}}{u_x^2}\right)=\frac{e\times 10\times {10}^{-1}}{m\times {\displaystyle \frac{e}{m}}}=1$

The angle of deviation of the path of electron as it comes out of the field is $\theta =45°$.