Answer
$\frac{q_3}{q} = -0.444$
Work Step by Step
In part (a) and part (b), we found that particle 3 is located at the point $(3.00~cm, 0)$
If the net electrostatic force on particle 1 is zero, then the forces on particle 1 due to the other two particles must be equal in magnitude and opposite in direction. Then particle 3 must have a negative charge.
We can find $\frac{q_3}{q}$:
$\frac{\vert q_1 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~r_1^2} = \frac{\vert q_1 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~r_2^2}$
$\frac{-q_3}{r_1^2} = \frac{q_2}{r_2^2}$
$q_3 = -\frac{r_1^2~q_2}{r_2^2}$
$q_3 = -\frac{(3.00~cm)^2~(4.00~q)}{(9.00~cm)^2}$
$\frac{q_3}{q} = -\frac{4.00}{9.00}$
$\frac{q_3}{q} = -0.444$
