The correct option is B
Total heat gained by ice is equal to the total heat lost by steam.
For ice to completely convert into water, heat required is $$m_1 L_f = 1 \times 80 = 80 cal$$
For steam to completely convert into water, heat released is $$m_2L_v = 1 \times 540 = 540\ cal$$
Hence, first 80 calories will not be enough for the steam to condense completely.
Now, to convert melted water to $$100^o C$$ from $$0^o C$$, heat required is $$m_1s(100-0)= 1 \times 1 \times 100 = 100 cal$$
So, total energy required to heat ice to water $$100^o C$$ is $$100+80=180\ cal$$.
Hence, even this amount of energy is not enough for the steam to condense completely. Hence, the final temperature of the mixture will be $$100^o C$$.
Note- finally the mixture will consist of both steam and water at $$100^o C$$.