#### Answer

$\alpha = 4$

#### Work Step by Step

Let $F_x$ be the magnitude of the force on the charge $-q$ (directed to the right) from the charge $\alpha q$. Let $F_y$ be the magnitude of the force on the charge $-q$ (directed upward) from the charge $q$.
Since the net force is directed at a $45^{\circ}$ angle, then $F_x = F_y$. We can find $\alpha$:
$F_x = F_y$
$\frac{k~q~\alpha q}{(2L)^2} = \frac{k~q~q}{L^2}$
$\frac{\alpha}{4} = 1$
$\alpha = 4$