Answer
$W_B = 1.375 \times 10^{28}$
Work Step by Step
Since configuration B has an these divisions of the molecules $n_1 = 0.6N$ and $n_2 = 0.4N$
The configuration of multiplicity for $N= 100$ is
$W_B = \frac{N!}{n_1!n_2!}$
$W_B = \frac{100!}{(0.6 \times 100)!(0.4 \times 100)!} $
$W_B = \frac{100!}{60!40!} $
$W_B = 1.375 \times 10^{28}$
