1. Given data

   • Mass of water: $m = 1\,\text{kg}$
   • Initial temperature: $20^\circ\text{C}$
   • Final state: fully evaporated at $100^\circ\text{C}$
   • Specific heat of water: $c = 4200\,\text{J/(kg}^\circ\text{C)}$
   • Latent heat of vaporization: $L = 2260\,\text{kJ/kg} = 2.26 \times 10^6\,\text{J/kg}$
   • Resistance of heating element: $R = 20\,\Omega$
   • RMS voltage: $V = 200\,\text{V}$

2. Electrical power supplied

   For a resistor, $P = \frac{V^2}{R}$ $P = \frac{(200)^2}{20} = \frac{40000}{20} = 2000\,\text{W}$

3. Heat required to raise temperature from $20^\circ\text{C}$ to $100^\circ\text{C}$ $Q_1 = mc\Delta T$ $Q_1 = 1 \times 4200 \times (100-20)$ $Q_1 = 4200 \times 80 = 336000\,\text{J}$

4. Heat required to evaporate 1 kg water at $100^\circ\text{C}$ $Q_2 = mL$ $Q_2 = 1 \times 2.26 \times 10^6 = 2.26 \times 10^6\,\text{J}$

5. Total heat required $Q = Q_1 + Q_2$ $Q = 336000 + 2.26 \times 10^6$ $Q = 2.596 \times 10^6\,\text{J}$

6. Time taken Using $t = \frac{Q}{P}$ $t = \frac{2.596 \times 10^6}{2000} = 1298\,\text{s}$

   Convert into minutes: $t = \frac{1298}{60} \approx 21.6\,\text{min}$

7. Closest option $t \approx 22\,\text{minutes}$

Therefore, the correct option is B.