Answer
$\frac{q_1}{q_2} = -\frac{5}{3}$
Work Step by Step
We can write a general expression for the electric potential energy of a system of charged particles:
$U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i~q_j}{r_{ij}}$
We can write an expression for the electric potential energy of the original two-particle system:
$U_i = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{d}$
We can write an expression for the electric potential energy of the three-particle system:
$U_f = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1~q_2}{d}+\frac{q_1~q_3}{2.5d}+\frac{q_2~q_3}{1.5d})$
It is given in the problem that $U_i = U_f$
We can find the ratio $\frac{q_1}{q_2}$:
$U_i = U_f$
$\frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{d} = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1~q_2}{d}+\frac{q_1~q_3}{2.5d}+\frac{q_2~q_3}{1.5d})$
$\frac{q_1~q_2}{d} =\frac{q_1~q_2}{d}+\frac{q_1~q_3}{2.5d}+\frac{q_2~q_3}{1.5d}$
$\frac{q_1~q_3}{2.5d} = -\frac{q_2~q_3}{1.5d}$
$\frac{q_1}{q_2} = -\frac{2.5}{1.5}$
$\frac{q_1}{q_2} = -\frac{5}{3}$
