Answer
The magnitude of the electric force on the $-q$ point charge due to the other two point charges is $\frac{k~q^2}{2d^2}$ and this force is directed toward the $+q$ point charge.
Work Step by Step
We can find the electric force on the $-q$ point charge due to the $+q$ point charge:
$F = \frac{k~(-q)~(+q)}{d^2}$
$F = -\frac{k~q^2}{d^2}$
The magnitude of the electric force on the $-q$ point charge due to the $+q$ point charge is $\frac{k~q^2}{d^2}$ and this force is directed toward the $+q$ point charge.
We can find the electric force on the $-q$ point charge due to the $+2q$ point charge:
$F = \frac{k~(-q)~(+2q)}{(2d)^2}$
$F = -\frac{k~q^2}{2d^2}$
The magnitude of the electric force on the $-q$ point charge due to the $+2q$ point charge is $\frac{k~q^2}{2d^2}$ and this force is directed toward the $+2q$ point charge.
We can find the vector sum of these two forces:
$\sum F = \frac{k~q^2}{d^2}-\frac{k~q^2}{2d^2} = \frac{k~q^2}{2d^2}$
The magnitude of the electric force on the $-q$ point charge due to the other two point charges is $\frac{k~q^2}{2d^2}$ and this force is directed toward the $+q$ point charge.
