Answer
$\frac{q_2}{q_1}= 9.0$
Work Step by Step
On the graph, we can see that when particle 3 is at $x = 2.0~cm$, the forces on particle 3 due to the other two particles have the same magnitude and they are in the opposite direction. Therefore, $q_1$ and $q_2$ have the same charge.
We can consider the point $x = 2.0~cm$ to find $\frac{q_2}{q_1}$:
$\frac{q_1~ q_3}{4\pi~\epsilon_0~r_1^2}= \frac{q_2~ q_3}{4\pi~\epsilon_0~r_2^2}$
$\frac{q_1}{r_1^2}= \frac{q_2}{r_2^2}$
$\frac{q_2}{q_1}= \frac{r_2^2}{r_1^2}$
$\frac{q_2}{q_1}= \frac{(6.0~cm)^2}{(2.0~cm)^2}$
$\frac{q_2}{q_1}= 9.0$
