Correct answer is B.
The die is thrown twice,
So the favourable outcomes that the sum appears to be 7 are
(1,6), (2,5), (3,4), (4,3), (5,2) and (6,1)
Out of these 2 appears twice,
So the probability that 2 appears at least once is:
\(=\frac{Favorable\,outcomes}{Total\,outcomes}\)
\(\Rightarrow \frac{2}{6}=\frac{1}{3}\)