Answer
$2.2\;\rm mm$
Work Step by Step
Since the electric field is uniform inside the capacitor, the force exerted on the proton 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
$$F=q_pE=m_pa$$
where $E=\eta/\epsilon_0$, so
$$a=\dfrac{q_p\eta}{m_p\epsilon_0}\tag 1$$
Now we know that the proton velocity $y$-component is constant since the electric force is on the $x$-direction as we see in the figure below.
So the distance traveled in the $x$-direction which is $x_1$ is given by
$$x_1=x_i+v_{ix}t+\frac{1}{2}at^2=0+0+\frac{1}{2}at^2$$
The proton starts at the origin, so $x_i=0$ and its initial $x$-velocity component is zero, so $v_{ix}=0$.
Plug $a$ from (1),
$$x_1=\frac{1}{2}\dfrac{q_p\eta}{m_p\epsilon_0}t^2\tag 2$$
Now we need to find the time $t$ of the trip from the start of the plate to the other end.
We know that $v_y$ is constant, so
$$v_y=\dfrac{y}{t}$$
where $y$ is the length of the plate and $v_y=v_0$ is the speed of the proton at the moment it enters between the two plates.
Hence,
$$t=\dfrac{y}{v_0}$$
Plug into (2),
$$x_1=\frac{1}{2}\dfrac{q_p\eta}{m_p\epsilon_0}\left[ \dfrac{y}{v_0}\right]^2 $$
Plug the known;
$$x_1=\frac{1}{2}\dfrac{(1.6\times 10^{-19})(1\times 10^{-6})}{(1.67\times 10^{-27})(8.85\times 10^{-12})}\left[ \dfrac{0.02}{1\times 10^6}\right]^2 $$
$$x_1=0.002165 \;\rm m=\color{red}{\bf 2.2}\;\rm mm$$
