Answer
$-1.92\times 10^{-13}~J~~$ of work was required to set up the system.
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 find the electric potential energy of this system of charged particles:
$U = \frac{1}{4\pi~\epsilon_0}~(\frac{-4q^2}{a}+\frac{2q^2}{\sqrt{2}~a})$
$U = \frac{q^2}{4\pi~\epsilon_0}~(\frac{-4}{a}+\frac{2}{\sqrt{2}~a})$
$U = \frac{(2.30\times 10^{-12}~C)^2}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~(\frac{-4}{0.64~m}+\frac{2}{\sqrt{2}~(0.64~m)})$
$U = -1.92\times 10^{-13}~J$
We can find the work that was required to set up the system:
$W = U = -1.92\times 10^{-13}~J$
$-1.92\times 10^{-13}~J~~$ of work was required to set up the system.
