B

Step 1: Calculate the energy released by steam when it condenses at 100 ºC.
Energy released = mass \times \text{latent heat of vaporization} = 1 \text{ kg} \times 2.26 \times 10^6 \text{ J/kg} = 2.26 \times 10^6 \text{ J}.

Step 2: Calculate the energy required to melt the ice at 0 ºC.
Energy required = mass \times \text{latent heat of fusion} = 1 \text{ kg} \times 3.36 \times 10^5 \text{ J/kg} = 3.36 \times 10^5 \text{ J}.

Step 3: Determine the energy balance.
The steam releases 2.26 \times 10^6 J, which is greater than the 3.36 \times 10^5 J required to melt the ice. Therefore, all the ice will melt, and the remaining energy will be used to increase the temperature of the resulting water.

Step 4: Calculate the remaining energy after melting the ice.
Remaining energy = 2.26 \times 10^6 \text{ J} - 3.36 \times 10^5 \text{ J} = 1.926 \times 10^6 \text{ J}.

Step 5: Calculate the temperature increase of the water.
The total mass of water after melting the ice is 2 kg. We use the formula: Q = mc\Delta T.
Here, Q = 1.926 \times 10^6 J, m = 2 kg, c (specific heat of water) = 4.18 \times 10^3 J/(kg K). Therefore, we have 1.926 \times 10^6 = 2 \times 4.18 \times 10^3 \Delta T.

Solving for \Delta T gives:
\Delta T = \frac{1.926 \times 10^6}{2 \times 4.18 \times 10^3} = \frac{1.926 \times 10^6}{8360} \approx 230 \text{ K} (this is above boiling point, so water will boil).

Therefore, some of the water will continue to vaporize until equilibrium is reached.

Conclusion: The final composition will include liquid water at 100 ºC and some steam. Thus, the answer is Composition: 1 kg water and some steam.