Answer
$D = 1.92~cm$
Work Step by Step
By symmetry, the vertical components of the forces on particle 1 due to particle 3 and particle 4 cancel out.
If the net electrostatic force on particle 1 due to the other particles is zero, then the force due to particle 2 and the sum of the horizontal components of the forces due to particle 3 and particle 4 must be equal in magnitude and opposite in direction.
We can equate the magnitudes to find $D$:
$2\times ~\frac{\vert q_1 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~(d/cos~\theta)^2}~cos~\theta = \frac{\vert q_1 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~(D+d)^2}$
$\frac{2~\vert~q_3 \vert}{d^2}~cos^3~\theta = \frac{\vert~q_2 \vert}{(D+d)^2}$
$\frac{2~\vert q_2 \vert}{d^2}~cos^3~\theta = \frac{5 \vert q_2 \vert}{(D+d)^2}$
$(D+d)^2 = \frac{5d^2}{2~cos^3~\theta}$
$(D+2.00)^2 = \frac{(5)(2.00)^2}{2~cos^3~30.0^{\circ}}$
$D^2+4.00D+4.00 = 15.396$
$D^2+4.00D-11.396 = 0$
We can use the quadratic formula:
$D = \frac{-4.00\pm \sqrt{(4.00)^2-(4)(1)(-11.396)}}{(2)(1)}$
$D = \frac{-4.00\pm \sqrt{(4.00)^2-(4)(1)(61.584)}}{2}$
$D = -5.92~cm, 1.92~cm$
Since $D$ is positive, the solution is $D = 1.92~cm$
