Answer
$W_B = 1$
Work Step by Step
System B has 5 particles (N). To know the least multiplicity configuration, all the particles must be in the same half of the box. This makes $n_1 = 5$ and $n_2 = 0$. Put these numbers into the multiplicity equation.
$W_B = \frac{N!}{n_1!n_2!}$
$W_B = \frac{5!}{5!0!} $
$W_B = 1$
