Answer
$6.4\times 10^{-6}\;\rm C/m^2$
Work Step by Step
Let's assume that the gravitational force is negligible and that the oil droplet is a point charge.
This means that the only force exerted on this negatively charged droplet is attraction electric force due to the electric field from the positively charged plate.
So according to Newton's second law,
$$F_{net}=ma$$
where $F_{net}=qE$
$$qE=ma$$
and we know that the electric field of a wide plane is given by $\eta/2\epsilon_0$.
$$q\dfrac{\eta}{2\epsilon_0}=ma$$
Solving for $\eta$ since the author asked to find the surface charge density of the plane.
$$\eta=\dfrac{2\epsilon_0ma}{q}$$
Note that the net charge of the droplet is $q=Ne$ where $N$ is the number of electrons added to it.
$$\eta=\dfrac{2\epsilon_0ma}{Ne}$$
We are not given the mass of the droplet but we can use the density law of $\rho=m/V$, so $m=\rho V$ and we can assume that the droplet is a perfect sphere, so $V=\frac{4\pi r^3}{3}$
$$\eta=\dfrac{ \frac{8}{3}\pi\epsilon_0\rho r^3 a}{Ne}\tag 1$$
To find the acceleration of the droplet, we can use the kinematic formula of
$$v_f^2=v_i^2+2ad=0^2+2ad$$
where it is released from rest,
so
$$a=\dfrac{v_f^2}{2d}$$
Plugging into (1),
$$\eta=\dfrac{ \frac{8}{3}\pi\epsilon_0\rho r^3 }{Ne}\cdot \dfrac{v_f^2}{2d}$$
$$\eta=\dfrac{ 4\pi\epsilon_0\rho r^3 v_f^2 }{3dNe} $$
Plug the known;
$$\eta=\dfrac{ 4\pi(8.85\times 10^{-12})(900)(0.5\times 10^{-6})^3 (3.5)^2 }{3(2\times 10^{-3})(20)(1.6\times 10^{-19})} $$
$$\eta=\color{red}{\bf 6.4\times 10^{-6}}\;\rm C/m^2$$
