A pure inductor of 0.30 H is joined to the 230 V, 50 Hz mains…

Step 1. The reactance. $X_L = 2\pi f L = 2\pi \times 50\ \text{Hz} \times 0.30\ \text{H} = 94.2\ \Omega$.

Step 2. The rms current. $I_{rms} = \dfrac{V_{rms}}{X_L} = \dfrac{230\ \text{V}}{94.2\ \Omega} = 2.44$ A.

Step 3. The peak current. $I_m = \sqrt{2}\, I_{rms} = 1.414 \times 2.44\ \text{A} = 3.45$ A.

Step 4. Half the current. The voltage stays the same, so the reactance must double. $X_L$ is in proportion to $f$, so the frequency must double: $100$ Hz.