Two charges 7 μ C and -2 μ C are placed at (-9,0,0) cm and (9,0,0)...

The total electrostatic energy $U$ of a system of two charges $q_1$ and $q_2$ in an external field is given by $U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}$.
Given external electric field $E = \frac{A}{r^2} \hat{r}$, the potential $V(r)$ is found by $V(r) = -\int_{\infty}^{r} E \cdot dr = -\int_{\infty}^{r} \frac{A}{r^2} dr = \frac{A}{r}$.
Parameters given:
$q_1 = 7 \mu C = 7 \times 10^{-6} C$ at $r_1 = 9 cm = 0.09 m$.
$q_2 = -2 \mu C = -2 \times 10^{-6} C$ at $r_2 = 9 cm = 0.09 m$.
Distance between charges $r_{12} = 9 - (-9) = 18 cm = 0.18 m$.
$A = 9 \times 10^5 N/C \cdot m^2$.
Calculating individual terms:
$q_1 V(r_1) = q_1 \frac{A}{r_1} = (7 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = 7 \times 10^{-6} \times 10^7 = 70 J$.
$q_2 V(r_2) = q_2 \frac{A}{r_2} = (-2 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = -2 \times 10^{-6} \times 10^7 = -20 J$.
Interaction energy $\frac{k q_1 q_2}{r_{12}} = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 J$.
Total energy $U = 70 - 20 - 0.7 = 49.3 J$.