Use the method of images. There is an image charge $-q$ along the $x$ axis at $x = -0.5 a$ and an image charge $-2q$ at $x = -1.5a$ along the $x$ axis. The force on the charge $q$ is therefore,
$\begin{align*}\mathbf{F} = \frac{q^2}{4 \pi \epsilon_0} \left(\frac{-1}{a^2} + \frac{-2}{a^2} + \frac{-2}{\left(2a \right)^2}...$
Therefore, answer (E) is correct.

Solution to 1992 Problem 10