Answer
$W_A = 1.01 \times 10^{29}$
Work Step by Step
Since configuration A has an equal division of the molecules between the two halves of the box, $n_1 = n_2 = N/2$
The configuration of multiplicity for $N= 100$ is
$W_A = \frac{N!}{n_1!n_2!}$
$W_A = \frac{100!}{50!50!} $
$W_A = 1.01 \times 10^{29}$
