Step 1: Use the formula for energy stored in a capacitor
\[
\text{Energy} = \frac{1}{2} C V^2
\]
Given:
- Capacitance \( C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F} \)
- Voltage \( V = 20 \, \text{V} \)
\[
\text{Energy} = \frac{1}{2} \times (10 \times 10^{-6}) \times (20)^2 = 0.01 \, \text{J}
\]
Answer: Therefore, the energy stored in the capacitor is \( 0.01 \, \text{J} \). So, the correct answer is option (1).