Recall the spectroscopic notation:
$\begin{align*}^{2 S +1} L\end{align*}$
So, in this problem, the orbital angular momentum quantum number is $0$ and the spin angular momentum quantum number is $1$ . Therefore, the total angular momentum quantum number can only have one possible value: $1$ . Therefore, answer (B) is correct.

Solution to 1992 Problem 31