Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to draw the force diagram of the particle.
We are told that the particle is falling at a constant speed (terminal speed), so the net force exerted on it is zero.
$$\sum F_y=F_D-mg-F_E=ma_y=m(0)=0$$
where $F_D$ is the drag force which is given by $F_D=6\pi \eta rv $, and $F_E$ is the electric force due to Earth's electric field.
$$6\pi \eta rv_{\rm term}-mg-qE=0$$
Solving for $v_{\rm term}$
$$\boxed{v_{\rm term}=\dfrac{mg+qE}{6\pi \eta r}}$$
$$\color{blue}{\bf [b]}$$
In the absence of Earth's electric field, the terminal speed of the particle is then given by
$$ v_{\rm term}=\dfrac{mg}{6\pi \eta r} $$
where $m=\rho V=\rho (4\pi r^3/3)$
$$ v_{\rm term}=\dfrac{4\rho \color{red}{\bf\not} \pi r^3g}{3(6\pi \eta \color{red}{\bf\not} r)} $$
$$ v_{\rm term}=\dfrac{2\rho r^2g}{9 \eta } $$
Plug the known;
$$ v_{\rm term}=\dfrac{2(2200) (0.5\times 10^{-6})^2(9.8)}{9 (1.8 \times 10^{-5} )} $$
$$v_{\rm term}=\color{red}{\bf 0.0665}\;\rm mm/s$$
$$\color{blue}{\bf [c]}$$
The electric field here is downward while the charge is negative, so it pulls the charge up, not down. This means that the electric force is upward.
Thus the boxed formula above would be
$$v_{\rm term}=\dfrac{mg-qE}{6\pi \eta r}$$
Here we just need to plug the known
$$v_{\rm term}=\dfrac{\frac{4}{3}\pi r^3 \rho g-qE}{6\pi \eta r}$$
$$v_{\rm term}=\dfrac{\frac{4}{3}\pi r^3 \rho g-Nq_eE}{6\pi \eta r}$$
Plug the known;
$$v_{\rm term}=\dfrac{\frac{4}{3}\pi (0.5\times 10^{-6})^3 (2200)(9.8)-(250)(1.6\times 10^{-19})(150)}{6\pi (1.8\times 10^{-5})(0.5\times 10^{-6})}$$
$$v_{\rm term}=\color{red}{\bf 0.031}\;\rm mm/s$$
