Answer
$Q = 55~\mu C$
Work Step by Step
If the initial acceleration of particle 3 is in the positive direction of the y axis, then the horizontal components of the forces on particle 3 due to the other two particles must cancel out.
We can find the magnitude of $Q_2$:
$\frac{\vert Q_1 \vert~\vert~Q_3 \vert}{4\pi~\epsilon_0~r_1^2}cos ~\theta_1 = \frac{\vert Q_2 \vert~\vert~Q_3 \vert}{4\pi~\epsilon_0~r_2^2}cos ~\theta_2$
$\frac{\vert Q_1 \vert}{r_1^2}cos ~\theta_1 = \frac{\vert Q_2 \vert}{r_2^2}cos ~\theta_2$
$\vert Q_2 \vert = \frac{r_2^2~cos ~\theta_1}{r_1^2~cos~\theta_2}~Q_1$
$\vert Q_2 \vert = \frac{(\sqrt{13}~cm)^2~(2.0/\sqrt{8.0})}{(\sqrt{8.0}~cm)^2~(3.0/\sqrt{13})}~40~\mu C$
$\vert Q_2 \vert = 55~\mu C$
If $Q_1$ and $Q_2$ have opposite signs, then the horizontal component of the net force on particle 3 can not be zero.
Since $Q_1$ and $Q_2$ must have the same sign, then $Q_2 = 55~\mu C$
That is, $~~Q = 55~\mu C$
