Answer
a) $ 3.56\times 10^3 \;\rm N/C$
b) $1\;\rm cm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the electric field is uniform inside the capacitor, the force exerted on the electron when it is moving between the plates is constant as well.
This means that the acceleration of the electron is constant.
According to Newton's second law, the acceleration of the electron is given by
$$F=q_e E=m_ea$$
$$E=\dfrac{m_ea}{q_e }\tag 1$$
Now we know that the electron velocity $x$-component is constant since the electric force is on the $y$-direction as we see in the given figure.
So the distance traveled in the $x$-direction which is $x_f$ is given by
$$x_f=v_{x}t$$
The electron's initial $x$-velocity component is $v_{ix}=v_0\cos45^\circ$.
$$x_f=v_0\cos45^\circ t$$
Now we need to find the time $t$ of the trip
$$t=\dfrac{x_f}{v_0\cos45^\circ}\tag 2$$
For $y$-direction,
$$y_f=y_i+v_{iy}t-\dfrac{1}{2}a_yt^2$$
where $a$ is negative since it pulls back the electron downward.
Plug the known;
$$0=0+v_{0}\sin45^\circ t-\dfrac{1}{2}a_yt^2$$
Thus,
$$a_y=\dfrac{2v_{0}\sin45^\circ }{t}$$
Plug from (2),
$$a_y=\dfrac{2v_{0}^2\sin45^\circ \cos45^\circ }{x_f}$$
$$a_y=\dfrac{ v_{0}^2\sin90^\circ }{x_f}$$
$$a_y=\dfrac{v_0^2}{x_f}\tag 3$$
Plug into (1)
$$E=\dfrac{m_ev_0^2}{x_fq_e } $$
Plugging the known;
$$E=\dfrac{(9.11\times 10^{-31})(5\times 10^6)^2}{(4\times 10^{-2})(1.6\times 10^{-19}) } $$
$$E=\color{red}{\bf 3.56\times 10^3}\;\rm N/C$$
$$\color{blue}{\bf [b]}$$
The smelling distance between the plates is the maximum height of the electron before it turns back toward the positive plate.
At this point $v_{fy}=0$ m/s,
$$v_{fy}^2=v_{iy}^2-2a_y d$$
where $d$ is the minimum distance between the two plates or the maximum height of the electron.
$$d=\dfrac{v_{fy}^2-v_{iy}^2}{-2a_y} =\dfrac{0^2-(v_0\sin45^\circ)^2}{-2a_y}$$
$$d=\dfrac{v_0^2}{4a_y}$$
Plug from (3),
$$d=\dfrac{v_0^2x_f}{4v_0^2}=\dfrac{x_f}{4}=\dfrac{4}{4}$$
$$d=\color{red}{\bf 1.0}\;\rm cm$$
