Solution :
$U= \large\frac{1}{4 \pi \in_0} \frac{q_1q_2}{r} $
$\quad= 9 \times 10^9 \times \large\frac{7 \times (-2) \times 10^{-12}}{0.18}$
$\qquad= -0.7\;J$
The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
$q_1V(r_1)+q_2V(r_2)=A $$\large\frac{\mu C}{0.09 \;m} +A \large\frac{-2 \mu C}{-0.09\;m}$
and the net electrostatic energy is
$\qquad= 70-20-0.7=49.3\;J$