I'll assume [tex]\mathrm B[/tex] is completely dominant over [tex]\mathrm b[/tex]. Recall the Hardy-Weinberg equations:
[tex]\begin{cases}p^2+2pq+q^2=1\\p+q=1\end{cases}[/tex]
where [tex]p[/tex] represents the allele frequency for brown fur, or the number of copies of the allele [tex]\mathrm B[/tex] within the population; and [tex]q[/tex] represents the allele frequency for white fur, or the number of copies of [tex]\mathrm b[/tex]. In the first equation, the squared terms refer to the frequencies of the corresponding homozygous individuals, while [tex]2pq[/tex] is the frequency of the heterozygotes.
We're told that 12 individual moles have white fur, so we know for sure that there are 12 homozygous recessive individuals, which means
[tex]q^2=\dfrac{12}{100}=\dfrac3{25}\implies q=\dfrac{\sqrt3}5\approx0.346[/tex]
from which it follows that
[tex]p=1-\dfrac{\sqrt3}5\approx0.654[/tex]
Over time, H-W equilibrium guarantees that the allele frequencies [tex](p,q)[/tex] do not change within the population. For example, suppose we denote the frequency of the [tex]\mathrm B[/tex] allele in generation [tex]n[/tex] by [tex]p_n[/tex]. Then
[tex]p_n={p_{n-1}}^2+\dfrac{2p_{n-1}q_{n-1}}2[/tex]
That is, the frequency [tex]\mathrm B[/tex] in the [tex]n[/tex]-th generation has to match the frequency of [tex]\mathrm B[/tex] attributed to the [tex]\mathrm{BB}[/tex] and half the [tex]\mathrm{Bb}[/tex] individuals of the previous generation.
[tex]p_n={p_{n-1}}^2+p_{n-1}q_{n-1}=p_{n-1}(p_{n-1}+q_{n-1})=p_{n-1}[/tex]
The same goes for [tex]q_n[/tex].
