Answer
$\frac{W_A}{W_B} = \frac{(N/2!)(N/2!)}{(N/3!)(N/3!)(N/3!)}$
Work Step by Step
Visualize $\frac{1}{2}$ of the molecules is in the right side of the box, and the other half is in the left side of the box. Then the multiplicity is
$ W_B= \frac{N!}{(N/2!)(N/2!)} $
and if $\frac{1}{3}$ of the molecules are in each third of the box, then the multiplicity is
$ W_A= \frac{N!}{(N/3!)(N/3!)(N/3!)} $
The ratio of $\frac{W_A}{W_B} $ is
$\frac{W_A}{W_B} = \frac{(N/2!)(N/2!)}{(N/3!)(N/3!)(N/3!)}$
