Answer
$q_1 = -20~nC$
Work Step by Step
Let $F_L$ be the magnitude of the force on the charge $q_2$ (directed to the left) from $q_1$. Let $F_R$ be the magnitude of the force on the charge $q_2$ (directed to the right) from the $5.0~nC$ charge.
Since $q_2$ is in static equilibrium, $F_L = F_R$. We can find the magnitude of the charge $q_1$:
$F_L = F_R$
$\frac{k~q_1~q_2}{(2r)^2} = \frac{k~(5.0~nC)~q_2}{r^2}$
$\frac{q_1}{4} = 5.0~nC$
$q_1 = 20~nC$
Since the charge on $q_1$ is negative, $q_1 = -20~nC$.
