1. **Problem statement:** Given a capacitor with capacitance $C_1 = 4 \times 10^{-6}$ F and charge $Q = 180 \times 10^{-6}$ C, the energy stored is $E_1 = 4.05 \times 10^{-3}$ J. We need to find the new capacitance $C_2$ when the charge remains the same but the energy stored changes to $E_2 = 4.05 \times 10^{-2}$ J. 2. **Formula used:** The energy stored in a capacitor is given by $$E = \frac{Q^2}{2C}$$ where $E$ is energy, $Q$ is charge, and $C$ is capacitance. 3. **Step 1: Express initial energy in terms of $C_1$ and $Q$:** $$E_1 = \frac{Q^2}{2C_1}$$ 4. **Step 2: Express new energy in terms of $C_2$ and $Q$:** $$E_2 = \frac{Q^2}{2C_2}$$ 5. **Step 3: Rearrange to find $C_2$:** $$C_2 = \frac{Q^2}{2E_2}$$ 6. **Step 4: Use the ratio of energies to relate $C_1$ and $C_2$:** $$\frac{E_2}{E_1} = \frac{\frac{Q^2}{2C_2}}{\frac{Q^2}{2C_1}} = \frac{C_1}{C_2}$$ 7. **Step 5: Solve for $C_2$:** $$C_2 = C_1 \times \frac{E_1}{E_2}$$ 8. **Step 6: Substitute values:** $$C_2 = 4 \times 10^{-6} \times \frac{4.05 \times 10^{-3}}{4.05 \times 10^{-2}} = 4 \times 10^{-6} \times \frac{1}{10} = 4 \times 10^{-7}$$ 9. **Answer:** The new capacitance is $4 \times 10^{-7}$ F, which corresponds to option (A).