Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, from the density law, that
$$\rho=\dfrac{m}{V}$$
So,
$$m=\rho V$$
the volume of the oil droplet is the volume of a sphere which is given by $\frac{4}{3}\pi R^3$, so
$$m=\frac{4}{3}\pi R^3\rho =\frac{4}{3}\pi (0.4\times 10^{-6})^3(885)$$
$$m=\color{red}{\bf 2.37\times 10^{-16}}\;\rm kg$$
$$\color{blue}{\bf [b]}$$
We know that the oil droplet is suspended in air between the two plates which means that the net force exerted on it is zero.
This also means that the electric force exerted on it is equal to its weight,
Thus,
$$F_E=F_G$$
$$Eq_{ \rm drop}=mg$$
where the electric field $E$ between the two plates is uniform and is given by $ ΔV/d $, hence,
$$q_{ \rm drop}=\dfrac{mg}{E}=\dfrac{mg d}{\Delta V}$$
Plug the known;
$$q_{ \rm drop} =\dfrac{(2.37\times 10^{-16})(9.8)(11\times 10^{-3})}{(20)}$$
$$q_{ \rm drop} =\color{red}{\bf1.28\times 10^{-18}}\;\rm C$$
$$\color{blue}{\bf [c]}$$
Since the positive plate is positioned upwards, the electric field is directed downwards. Consequently, if the droplet were positively charged, both the gravitational force and the electric force would act in the same downward direction, causing the droplet to fall.
Hence, the droplet must be negatively charged, resulting in an upward electric force.
To determine the number of electrons, we can divide the droplet's charge by the charge of a single electron.
$$N=\dfrac{q_{ \rm drop} }{e}=\dfrac{1.28\times 10^{-18}}{1.6\times 10^{-19}}=7.94$$
$$N\approx\color{red}{\bf 8}\;\rm e^-$$
