Answer
$x = 6.05~cm$
Work Step by Step
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.
Therefore, initially, the net force is directed upward.
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.
In part (a), we found that particle 1 needed to be placed at $x = -6.05~cm$
To rotate $F_{net}$ back to its original position, the horizontal component of the net force must be zero.
Then, by symmetry, particle 3 must be placed at $x = 6.05~cm$
