Answer
$x = -6.05~cm$
Work Step by Step
Let $d = 5.00~cm$
We can find an expression for the magnitude of the vertical component of the net force on particle 5:
$F_y = \frac{e^2}{4\pi~\epsilon_0~d^2}-\frac{e^2}{4\pi~\epsilon_0~4d^2}$
$F_y = \frac{3}{4}(\frac{e^2}{4\pi~\epsilon_0~d^2})$
Note that the vertical component of the net force on particle 5 is directed upward.
Initially, by symmetry, the horizontal component of the net force on particle 5 is zero.
If the direction of the net force on particle 5 is rotated $30^{\circ}$ counterclockwise, then the direction of the net force on particle 5 will be $30^{\circ}$ to the left of the positive y axis.
We can find the required magnitude of the horizontal component of the net force:
$\frac{F_x}{F_y} = tan~30^{\circ}$
$F_x = F_y~tan~30^{\circ}$
$F_x = (\frac{3}{4})(\frac{e^2}{4\pi~\epsilon_0~d^2})~tan~30^{\circ}$
$F_x = 0.433~\frac{e^2}{4\pi~\epsilon_0~d^2}$
We can find $x$, the required distance of particle 1 from the origin:
$F_x = \frac{e^2}{4\pi~\epsilon_0~x^2}-\frac{e^2}{4\pi~\epsilon_0~4d^2} = 0.433~\frac{e^2}{4\pi~\epsilon_0~d^2}$
$(\frac{1}{x^2}-\frac{1}{4d^2})~\frac{e^2}{4\pi~\epsilon_0} = \frac{0.433}{d^2}~\frac{e^2}{4\pi~\epsilon_0}$
$\frac{1}{x^2}= \frac{0.433}{d^2}+\frac{1}{4d^2}$
$\frac{1}{x^2}= \frac{1.732}{4d^2}+\frac{1}{4d^2}$
$\frac{1}{x^2}= \frac{2.732}{4d^2}$
$x^2= \frac{4d^2}{2.732}$
$x= -\frac{2d}{\sqrt{2.732}}$
$x= -\frac{(2)(5.00~cm)}{\sqrt{2.732}}$
$x = -6.05~cm$
Since particle 1 is to the left of the origin, $x = -6.05~cm$
