The primary coil of a transformer has a resistance of 3.0 Ω and, at 50 Hz, an inductive reactance…

Step 1. The impedance on AC. $Z = \sqrt{R^2 + X_L^2} = \sqrt{(3.0\ \Omega)^2 + (345\ \Omega)^2} = 345\ \Omega$. The reactance decides.

Step 2. The current on AC. $I = \dfrac{V}{Z} = \dfrac{230\ \text{V}}{345\ \Omega} = 0.667$ A.

Step 3. The current on DC. The current is steady, so the coil has no reactance. Only $R$ limits the current: $I = \dfrac{V}{R} = \dfrac{230\ \text{V}}{3.0\ \Omega} = 76.7$ A.

Step 4. The heating power on DC. $P = I^2R = (76.7\ \text{A})^2 \times 3.0\ \Omega = 17\,600\ \text{W} = 17.6$ kW.