Answer
$\frac{q_1}{q_2} = -25.0$
Work Step by Step
If the net electrostatic force on particle 3 is zero, then the forces on particle 3 due to the other two particles must be equal in magnitude and opposite in direction.
We can find $\frac{\vert q_1 \vert}{\vert q_2 \vert}$:
$\frac{\vert q_1 \vert~Q}{4\pi~\epsilon_0~r_1^2}= \frac{\vert q_2 \vert~Q}{4\pi~\epsilon_0~r_2^2}$
$\frac{\vert q_1 \vert}{r_1^2}= \frac{\vert q_2 \vert}{r_2^2}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert}= \frac{r_1^2}{r_2^2}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert}= \frac{(2.50~a)^2}{(0.500~a)^2}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert}= \frac{6.25}{0.250}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert}= 25.0$
Since the forces on particle 3 due to the other two particles must be opposite in direction, $q_1$ and $q_2$ must have the opposite signs.
Then:
$\frac{q_1}{q_2} = -\frac{\vert q_1 \vert}{\vert q_2 \vert}= -25.0$
