Answer
The x coordinate of particle 3 is $~~x = 3.00~cm$
Work Step by Step
If the net electrostatic force on particle 3 due to the other two particles is zero, then the force vectors due to the other two particles must point in opposite directions. This is only possible if particle 3 is located on the x-axis.
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 3 must be located between the other two particles.
We can find $x$, the position of particle 3 on the x axis:
$\frac{\vert q_1 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~r_1^2} = \frac{\vert q_2 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~r_2^2}$
$\frac{q_1}{x^2} = \frac{q_2}{(9.00-x)^2}$
$\frac{q_1}{x^2} = \frac{4.00~q_1}{(9.00-x)^2}$
$(9.00-x)^2 = 4.00~x^2$
$x^2-18.0~x+81.0 = 4.00~x^2$
$3.00~x^2+18.0~x-81.0 = 0$
$x^2+6.00~x-27.0 = 0$
$(x-3.00)(x+9.00) = 0$
$x = 3.00~cm, -9.00~cm$
Since x must be positive, the solution is $x = 3.00~cm$
The x coordinate of particle 3 is $~~x = 3.00~cm$
