Solution:
The answer is (B). If we fix a boy, there are $5!$ possible permutations of the other 5 guests; among these, the ones in which boys and girls alternate are $3 \cdot 2 \cdot 2$, because the person to the right of the fixed boy can be any of the 3 girls, the person further to the right can be one of the two remaining boys, then there must be one of the other two girls, and the last two seats must necessarily be occupied by the last boy and the last girl.
The solution is therefore the number of favorable cases over the number of possible cases, that is $\frac{3 \cdot 2 \cdot 2}{5!}=\frac{1}{10}$.