Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need the electron to turn back to the same plate, so this plate must be positively charged.
That's because we need the electric field force to point the electron back toward the same plate.
$$\color{blue}{\bf [b]}$$
As the author told us, we need to be careful on choosing our coordinates. So we chose the $x$-coordinate to be parallel to the plates and the $y$-coordinate to be perpendicular to the plates.
Now we have a $y$-component acceleration and zero $x$-component acceleration as if the plates were horizontal and the electron is fired between them [see the previous problem].
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 proton is constant.
According to Newton's second law, the acceleration of the electron is given by
$$F=q_eE=m_ea$$
Thus,
$$a_y=\dfrac{q_eE}{m_e}\tag 1$$
Now we know that the proton velocity $x$-component is constant since the electric force is on the $y$-direction.
The distance traveled in the $x$-direction which is $x_1$ is 1 cm.
So,
$$\Delta x=v_xt$$
where $v_x=v_0\cos\theta$
Hence,
$$t=\dfrac{\Delta x}{v_0\cos\theta}\tag 2$$
Using the kinematic formula for $y$-direction,
$$\Delta y=v_{iy}t-\frac{1}{2}a_yt^2 $$
The electron starts at the origin and ends at the origin as well, so $\Delta y=0$ and its initial $y$-velocity component is $v_0\sin\theta$.
$$0= v_0\sin\theta t-\frac{1}{2}a_yt^2 $$
Hence,
$$ 2v_0\sin\theta =a_yt $$
Plug from (1) and (2)
$$ 2v_0\sin\theta =\dfrac{q_eE}{m_e} \dfrac{\Delta x}{v_0\cos\theta}$$
Solving for $E$,
$$E=\dfrac{2\sin\theta\cos\theta m_e v_0^2}{q_e\Delta x}$$
$$E=\dfrac{\sin(2\theta) m_e v_0^2}{q_e\Delta x}$$
where from the geometry of the figure below, we can see that $\theta=45^\circ\rightarrow\sin(2\theta)=\sin90^\circ=1 $.
Thus,
$$E=\dfrac{ m_e v_0^2}{q_e\Delta x}$$
we can find $v_0$ from the kinetic energy, $K=\frac{1}{2}m_ev_0^2$, so that $v_0^2=2K/m_e$.
$$E=\dfrac{ 2m_e K}{m_eq_e\Delta x}$$
$$E=\dfrac{ 2 K}{ q_e\Delta x}$$
Plug the known;
$$E=\dfrac{2 (3\times 10^{-17}) }{(1.6\times 10^{-19})(0.01)}$$
$$E=\color{red}{\bf 3.75\times 10^4}\;\rm N/C$$
$$\color{blue}{\bf [c]}$$
The minimum separation distance between the plate equals the maximum height of the electric in $y$-direction.
This happens when the $y$-component velocity of the electron is zero.
$$v_y^2=v_{iy}^2-2a_y d$$
Hence,
$$d_{min}=\dfrac{v_y^2-v_{iy}^2}{-2a_y }=\dfrac{0-v_{iy}^2}{-2a_y }$$
Plug from (1),
$$d_{min}= \dfrac{ m_ev_0^2(\sin45^\circ)^2}{ 2 q_eE}$$
where $v_0^2=2K/m_e$
$$d_{min}= \dfrac{ 2 K(\sin45^\circ)^2}{ 2 q_eE}$$
$$d_{min}= \dfrac{ 2(3\times 10^{-17}) (\sin45^\circ)^2}{ 2 (1.6\times 10^{-19})(3.75\times 10^4)}=0.0025\;\rm m$$
$$d_{min}=\color{red}{\bf 2.5}\;\rm mm$$
