Answer
$\approx 18.5\;\rm cm$
Work Step by Step
This problem is similar to situations involving free-fall acceleration, such as when you launch a baseball vertically upward and need to measure its final height before it falls back to the ground.
However, in this scenario, the force acting on the proton is an electric force resulting from the presence of an electric field, rather than the gravitational force.
Applying Newton's second law,
$$-F= m_pa=q_pE$$
Hence,
$$a_p=\dfrac{-q_pE}{m_p}\tag 1$$
where $_p$ refers to the proton.
Now we know the acceleration exerted on the proton that will pull it back to the negative plate. The negative charge is due to the force direction that opposes the proton's initial speed direction.
Recall the kinematic formula of
$$v_f^2=v_i^2+2ad$$
where the final speed of the proton at its turning point is zero.
Solving for $d$,
$$d=\dfrac{0^2-v_i^2}{2a_p}$$
Plugging from (1),
$$d=\dfrac{ -v_i^2}{2 }\cdot \dfrac{m_p}{-q_pE}=\dfrac{v_i^2m_p}{2q_pE}$$
Recalling that the electric field of a charged plane is $E=\eta/2\epsilon_0$,
$$d= \dfrac{\color{red}{\bf\not} 2\epsilon_0v_i^2m_p}{\color{red}{\bf\not} 2q_p\eta}$$
Plug the known;
$$d= \dfrac{ (8.85\times 10^{-12})(2\times 10^6)^2 (1.67\times 10^{-27})}{(1.6\times 10^{-19})(2\times 10^{-6})}=0.185\;\rm m$$
$$d\approx \color{red}{\bf 18.5}\;\rm cm$$
