The electrostatic energy stored in a capacitor can be calculated using the formula:

$$ U = \frac{1}{2} C V^2 $$

Where:

• $ U $ is the stored energy in joules (J)
• $ C $ is the capacitance in farads (F)
• $ V $ is the voltage in volts (V)

Given:

• $ C = 12 \mathrm{~pF} $ (which is $ 12 \times 10^{-12} $ F)
• $ V = 50 \mathrm{~V} $

Substitute the values into the formula:

$$ U = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2 $$

Calculate the values inside the parentheses first:

$ 50^2 = 2500 $

Now substitute this back into the equation:

$$ U = \frac{1}{2} \times 12 \times 10^{-12} \times 2500 $$

Perform the multiplication:

$$ U = \frac{1}{2} \times 30000 \times 10^{-12} $$

Which simplifies to:

$$ U = 15000 \times 10^{-12} $$

Convert this to nanojoules (nJ) by recognizing that $$ 1 \mathrm{~nJ} = 10^{-9} \mathrm{~J} $$:

$ U = 15 \mathrm{~nJ} $

Thus, the electrostatic energy stored in the capacitor is:

Option A: 15 nJ