Answer
$\theta = 51.0^{\circ}$
Work Step by Step
Since $q \leq 5e$, the charge of particles 3 and 4 can be $-e, -2e, -3e, -4e,$ or $-5e$
By symmetry, the vertical components of the forces on particle 2 due to particle 3 and particle 4 cancel out.
If the net electrostatic force on particle 2 due to the other particles is zero, then the force due to particle 1 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.
The second smallest angle $\theta$ will occur when the magnitude of particles 3 and 4 is the second smallest possible, which is $q = 2e$
We can equate the magnitudes to find $\theta$:
$2\times ~\frac{\vert q_2 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~(R/cos~\theta)^2}~cos~\theta = \frac{\vert q_1 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~R^2}$
$\frac{(2)~(2e)}{R^2}~cos^3~\theta = \frac{e}{R^2}$
$cos^3~\theta = \frac{1}{4}$
$cos~\theta = \sqrt[3] {\frac{1}{4}}$
$\theta = cos^{-1}~(\sqrt[3] {\frac{1}{4}})$
$\theta = 51.0^{\circ}$
