Answer
a) $3.6\times 10^6\;\rm N/C$
b) $8.3\times 10^5\;\rm m/s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The distance between the two disks is too small which means that we can consider this system as a capacitor.
We know that the electric field inside a capacitor is given by
$$E=\dfrac{\eta}{\epsilon_0}$$
where $\eta=Q/A$ where $A$ is the surface area of the capacitor plate which is an area of a circle $\pi R^2$.
Thus,
$$E=\dfrac{Q}{\pi R^2\epsilon_0}$$
Plug the known;
$$E=\dfrac{10\times 10^{-9}}{\pi (1\times 10^{-2})^2(8.85\times 10^{-12})}$$
$$E=\color{red}{\bf 3.6\times 10^6}\;\rm N/C$$
$$\color{blue}{\bf [a]}$$
For the proton to just reach the positive plate when released from the negative plate, we need its final speed at the positive plate to be zero.
So, we need to find the force exerted on the proton by the electric field that would fire it from the positive plate from rest toward the negative plate. Its speed at the negative plate is the speed we need to fire it to travel in the opposite direction to reach the positive plate at zero speed.
Thus, according to Newton's second law, its acceleration is given by
$$F=q_pE=ma$$
Hence,
$$a=\dfrac{q_pE}{m}\tag 1$$
Now we can use the kinematic formula of
$$v^2=v_i^2+2ad$$
Plugging from (1), and recalling that we assumed that the proton is released from rest from the positive plate.
$$v^2=0^2+2d\dfrac{q_pE}{m}$$
Hence,
$$v=\sqrt{\dfrac{2dq_pE}{m}}$$
Plugging the known;
$$v=\sqrt{\dfrac{2 (1\times 10^{-3})(1.6\times 10^{-19})(3.6\times 10^6)}{(1.67\times 10^{-27})}}$$
$$v=\color{red}{\bf 8.3\times 10^5}\;\rm m/s$$
Thus the proton must be released at this speed from the negative plate to barely reach the positive plate.
