Answer
$q = 0.68~nC$
Work Step by Step
The force due to the $q$ charge must be equal and opposite to the sum of the vertical components of the forces due to the $2.0~nC$ charges.
We can find the angle $\theta$ below the horizontal of the direction of the force on the $1.0~nC$ due to the $2.0~nC$ charges:
$tan~\theta = \frac{3.0~cm}{2.0~cm}$
$\theta = tan^{-1}~(1.5)$
$\theta = 56.3^{\circ}$
We can find $q$:
$\frac{k~q~(1.0~nC)}{(2.0~cm)^2} = 2\times \frac{k~(2.0~nC)~(1.0~nC)}{(\sqrt{13}~cm)^2}~cos~56.3^{\circ}$
$q = \frac{(2)(2.0~nC)(2.0~cm)^2}{(\sqrt{13}~cm)^2}~cos~56.3^{\circ}$
$q = 0.68~nC$
