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, $q_1 V(r_1)+q_e V(r_2) =A \large\frac{7 \mu C}{0.09 m} +A \large\frac{-2 \mu C}{0.09 m}$

and the net electrostatic energy is

$q_1V(r_1)+q_2V(r_2)+\large\frac{q_1q_2}{4 \pi \in_0 r_{12} } $$=A \large\frac{7 \mu C}{0.09 m} + A \large\frac{-2 \mu C}{0.09 \;m}$$-0.7 J$

$\qquad= 70-20-0.7 =49.3 J$

Hence C is the correct answer.