Answer
$\frac{q_1}{q_2} = -2.25$
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. Then particle 1 and particle 2 must have opposite charges.
We can equate the magnitude of the forces to find $\frac{\vert q_1 \vert}{\vert q_2 \vert}$:
$\frac{\vert q_1 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~(L_{13})^2} = \frac{\vert q_2 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~(L_{23})^2}$
$\frac{\vert q_1 \vert}{(3.00~L_{12})^2} = \frac{\vert q_2 \vert}{(2.00~L_{12})^2}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert} = \frac{(3.00~L_{12})^2}{(2.00~L_{12})^2}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert} = \frac{9.00}{4.00}$
$\frac{\vert q_1 \vert}{\vert q_2 \vert} = 2.25$
Since particle 1 and particle 2 have opposite charges, $~~\frac{q_1}{q_2} = -2.25$
