$$ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} $$

Here,

$$ q_1 = 7 \times 10^{-6} \, C $$

$$ q_2 = -4 \times 10^{-6} \, C $$

The separation between the charges is along the x-axis from $$ -7 \, \text{cm} $$ to $$ 7 \, \text{cm} $$, so

$$ r = 0.14 \, m $$

The Coulomb constant is approximated as

$$ \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2} $$

Substitute these values into the equation:

$$ U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{0.14} $$

Calculate the product of the charges:

$$ (7 \times 10^{-6})(-4 \times 10^{-6}) = -28 \times 10^{-12} \, C^2 $$

Now, divide by the distance:

$$ \frac{-28 \times 10^{-12}}{0.14} = -2 \times 10^{-10} \, C^2/m $$

Finally, multiply by the Coulomb constant:

$$ U = 9 \times 10^9 \times (-2 \times 10^{-10}) = -1.8 \, J $$

Thus, the electrostatic potential energy of the configuration is:

$$ -1.8 \, J $$

This corresponds to Option D.