A

Step 1: Analyze the circuit at t = 0.
The capacitor is initially charged and then switches to the inductor. Let the initial voltage across the capacitor be V.
Step 2: The energy stored in the capacitor (E_c) is given by \(E_c = \frac{1}{2} C V^2\).
Step 3: After the switch is placed at (2), the inductor will start to receive energy. The current in the inductor (i) can be expressed over time \(i(t) = \frac{V}{R}(1 - e^{-\frac{R}{L}t})\).
Step 4: The energy in the inductor (E_l) is given by \(E_l = \frac{1}{2} L i^2 = \frac{1}{2} L \left(\frac{V}{R}(1 - e^{-\frac{R}{L}t})\right)^2\).
Step 5: Set E_c = E_l and solve for t. This gives:\
\(\frac{1}{2} C V^2 = \frac{1}{2} L \left(\frac{V}{R}(1 - e^{-\frac{R}{L}t})\right)^2\)
Step 6: Further simplification leads to \(C = \frac{L}{R^2}(1 - e^{-\frac{R}{L}t})^2\).
Step 7: Solving this equation will give the specific time at which the energies are equal. Depending on the values of R, C, and L provided in the question, t should be calculated.
Therefore, the time when the energy in both the capacitor and inductor will be the same for the first time can be derived from this process.