To calculate the electrostatic potential energy (U) of the system of charges, we use the formula for the potential energy between two point charges:

$$U = k \frac{q_1 q_2}{r}$$

where:

• U = potential energy between the charges
• k = Coulomb's constant = 8.99 \times 10^9 \frac{N m^2}{C^2}
• q_1 and q_2 = magnitudes of the charges
• r = distance between the charges

Given are charges:

• Charge 1 (C1) = +q = +10^{-7} C
• Charge 2 (C2) = +2q = +2 \times 10^{-7} C
• Charge 3 (C3) = -4q = -4 \times 10^{-7} C

Distance between each pair of charges is 10 cm (0.1 m). Now calculate the potential energy for each pair:

1. Between C1 and C2:

$$U_{12} = k \frac{(10^{-7})(2 \times 10^{-7})}{0.1} = 8.99 \times 10^9 \frac{(10^{-7})(2 \times 10^{-7})}{0.1}$$

Calculating:

$$U_{12} = 8.99 \times 10^9 \cdot \frac{2 \times 10^{-14}}{0.1} = 1.798 \times 10^{-3} J$$

2. Between C1 and C3:

$$U_{13} = k \frac{(10^{-7})(-4 \times 10^{-7})}{0.1} = -3.596 \times 10^{-3} J$$

3. Between C2 and C3:

$$U_{23} = k \frac{(2 \times 10^{-7})(-4 \times 10^{-7})}{0.1} = -7.192 \times 10^{-3} J$$

Now sum the potential energies:

$$U = U_{12} + U_{13} + U_{23}$$

Substituting values:

$$U = 1.798 \times 10^{-3} - 3.596 \times 10^{-3} - 7.192 \times 10^{-3}$$

Calculating:

$$U = 1.798 - 3.596 - 7.192 = -9.99 \times 10^{-3} J$$

Thus, the electrostatic potential energy of the system is:

Answer: -9.99 mJ

Therefore, the correct answer is Option A.