Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the droplet is suspended motionless, the net force exerted on it is zero.
$$\sum F=F_e-mg=0$$
So,
$$qE_0=mg$$
Hence,
$$\boxed{q=\dfrac{mg}{E_0}}\tag 1$$
$$\color{blue}{\bf [b]}$$
The droplet reaches the terminal speed when the net force exerted on it is zero.
$$\sum F=D-mg=0$$
So,
$$bv_{\rm terminal}=mg$$
$$\boxed{ v_{\rm terminal}=\dfrac{mg}{b}}\tag 2$$
$$\color{blue}{\bf [c]}$$
We are given that the drag force is given by
$$D=-6\pi \eta r v=-bv$$
So,
$$b=6\pi \eta r$$
Plug into (2);
$$ v_{\rm terminal}=\dfrac{mg}{6\pi \eta r}$$
where $m=\rho V$; and $V=\frac{4}{3}\pi r^3$
$$ v_{\rm terminal}=\dfrac{\frac{4}{3}\color{red}{\bf\not} \pi r^{{\color{red}{\bf\not} 3}^2} \rho g}{6\color{red}{\bf\not} \pi \eta\color{red}{\bf\not} r}$$
$$ v_{\rm terminal}=\dfrac{2 r^2\rho g}{9 \eta }$$
Thus the radius of the droplet is given by
$$ r^2=\dfrac{9 \eta v_{\rm terminal}}{2\rho g}$$
$$ \boxed{r =\sqrt{\dfrac{9 \eta v_{\rm terminal}}{2\rho g}}}\tag 3$$
$$\color{blue}{\bf [d]}$$
To find the charge, we need to use (1),
$$q=\dfrac{mg}{E_0}$$
where $m=\rho V=\frac{4}{3}\pi r^3 \rho $, and $E_0=\Delta V/d$
$$q=\dfrac{\frac{4}{3}\pi r^3 \rho g d}{\Delta V }$$
Plug $r$ from (3);
$$q=\dfrac{\frac{4}{3}\pi \rho g d}{\Delta V }\left[\sqrt{\dfrac{9 \eta v_{\rm terminal}}{2\rho g}}\right]^3$$
where $v_{\rm terminal}=\Delta y/\Delta t$;
$$q=\dfrac{\frac{4}{3}\pi \rho g d}{\Delta V }\left[\sqrt{\dfrac{9 \eta \dfrac{\Delta y}{\Delta t}}{2\rho g}}\right]^3$$
$$q=\dfrac{\frac{4}{3}\pi \rho g d}{\Delta V }\left[{\dfrac{9 \eta_{\rm air} \dfrac{\Delta y}{\Delta t}}{2\rho g}}\right]^{3/2}$$
Plug the known;
$$q=\dfrac{\frac{4}{3}\pi (860) (9.8)(0.01)}{(1177) }\left[{\dfrac{9 (1.83\times 10^{-5})\dfrac{3\times 10^{-3}}{7.33}}{2 (860) (9.8)}}\right]^{3/2}$$
$$q=\color{red}{\bf 2.4\times 10^{-18}}\;\rm C$$
$$\color{blue}{\bf [e]}$$
$$\dfrac{q}{e}=\dfrac{2.4\times 10^{-18}}{1.6\times 10^{-19}}=15$$
$$
\boxed{q=\color{red}{\bf 15}\;e }$$
