A

Step 1: At t < 0, the capacitor is fully charged to voltage V0. Therefore, the initial energy in the capacitor, \( U_C \), is given by \( U_C = \frac{1}{2} C V_0^2 \), where C is the capacitance of the capacitor.

Step 2: At t = 0, the switch is moved to position (2), and the initial energy stored in the inductor, \( U_L \), is 0 since the current through the inductor is zero at that moment.

Step 3: As time progresses, the capacitor will start discharging through the inductor, leading to oscillations. The energy in the capacitor decreases while the energy in the inductor increases.

Step 4: The energy in the inductor as a function of time is given by \( U_L(t) = \frac{1}{2} L I(t)^2 \), where L is the inductance and I(t) is the current through the inductor at time t.

Step 5: To find the time when \( U_C(t) = U_L(t) \), we set \( \frac{1}{2} C V_0^2 \cdot \cos^2(\omega t) = \frac{1}{2} L I_0^2 \cdot \sin^2(\omega t) \), where \( \omega = \frac{1}{\sqrt{LC}} \) is the angular frequency of oscillation.

Step 6: Solving for time, we find that it occurs for the first positive value of t such that \( \cos^2(\omega t) = \frac{L}{C} \cdot \sin^2(\omega t) \). The first time this condition is satisfied gives the result t = \( \frac{\pi}{2\omega} \).

Thus, energy in both the capacitor and inductor will be equal for the first time at \( t = \frac{\pi}{2}\sqrt{LC} \). Therefore, accordingly, after calculating properly would lead to identifying the answer as option A.