Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that the electron accelerates from rest, so $v_i=0$ and $K_i=0$. And then its final speed is given by the kinetic energy it gains
$$K_f=\frac{1}{2}m v_f^2$$
Hence,
$$v_f^2=\dfrac{2K_f}{m }\tag 1$$
This is the final velocity of the electron and if we assume that it is speeding up at a constant rate, $a=$constant. Hence, we can use the kinematic formula of
$$v_f^2=v_i^2+2ad=0+2ad$$
Thus, its acceleration
$$a=\dfrac{v_f^2}{2d}$$
Plugging from (1);
$$a=\dfrac{2K_f}{2md}=\dfrac{K_f}{md}$$
Plugging the known;
$$a =\dfrac{(2\times 10^{-18})}{(9.11\times 10^{-31})(2\times 10^{-6})}$$
$$a=\color{red}{\bf 1.1 \times 10^{18}}\;\rm m/s^2$$
$$\color{blue}{\bf [b]}$$
According to Newton's second law, the force exerted on the electron to give it this acceleration is given by
$$ F=ma=(9.11\times 10^{-31})(1.1 \times 10^{18})$$
$$F=\color{red}{\bf 1.0 \times 10^{-12}}\;\rm N$$
$$\color{blue}{\bf [c]}$$
We know that the electric field is given by
$$E=\dfrac{F}{q}=\dfrac{1.0 \times 10^{-12}}{1.6\times 10^{-19}}$$
$$E=\color{red}{\bf 6.25\times 10^6}\;\rm N/C$$
$$\color{blue}{\bf [d]}$$
We need the point charge to give the electron a force as we calculated in part (b).
So according to Coulomb's law,
$$F=\dfrac{kq_1q_2}{r^2}$$
where $q_1=q_e$, and $q_2 $ is the charge of the point charge.
$$F=\dfrac{kq_eq_2}{r^2}$$
Hence,
$$q_2=\dfrac{Fr^2}{kq_e}$$
Noting that we are neglecting the signs of the charges and focusing on the charge magnitude.
Plug the known;
$$q_2=\dfrac{ (1\times 10^{-12})(1\times 10^{-2})^2}{(8.99\times 10^{9})(1.6\times 10^{-19})}$$
$$q_2=6.95\times 10^{-8}\;\rm C\approx\color{red}{ \bf 70}\;nC$$