A source of 12 V rms has a frequency of 50 Hz…

Step 1. The coil at $400$ Hz. The frequency is $8$ times larger, so $X_L = 8 \times 40\ \Omega = 320\ \Omega$. The current is $\dfrac{12\ \text{V}}{320\ \Omega} = 0.0375\ \text{A} = 37.5$ mA.

Step 2. The capacitor at $400$ Hz. Now $X_C = \dfrac{40\ \Omega}{8} = 5.0\ \Omega$. The current is $\dfrac{12\ \text{V}}{5.0\ \Omega} = 2.4$ A.

Step 3. The frequency for $0.60$ A. The reactance needed is $\dfrac{12\ \text{V}}{0.60\ \text{A}} = 20\ \Omega$. That is half of $40\ \Omega$, so the frequency is twice $50$ Hz: $100$ Hz.