Answer
$W_A = 3$
Work Step by Step
System A has 3 particles (N). To know the $greatest$ multiplicity configuration, all the particles must be distributed equally in the same half of the box.
If the particle is odd number, put an extra particle in one of the boxes. This makes $n_1 = 2$ and $n_2 = 1$. Put these numbers into the multiplicity equation.
$W_A = \frac{N!}{n_1!n_2!}$
$W_A = \frac{3!}{2!1!} $
$W_A = 3$
