#### Answer

$F_x = \frac{-2~a~k~q~Q}{(a^2+y^2)^{3/2}}$

#### Work Step by Step

We can find the distance $r$ from the charge $q$ to each of the other charges.
$r = \sqrt{a^2+y^2}$
By symmetry, the vertical component of each force on the charge $q$ cancels out. The net force on the charge $q$ is the sum of the horizontal component of each force. We can find the magnitude of the horizontal component of the force on the charge $q$.
$F = 2\times \frac{k~q~Q}{r^2}~cos(\theta)$
$F = 2\times \frac{k~q~Q}{(\sqrt{a^2+y^2})^2}~(\frac{a}{\sqrt{a^2+y^2}})$
$F = \frac{2~a~k~q~Q}{(a^2+y^2)^{3/2}}$
Since the x-component of the force is directed to the left, the x-component of the force is negative.
$F_x = \frac{-2~a~k~q~Q}{(a^2+y^2)^{3/2}}$