help with hardy weinberg problem?

Fr(BB)=0.36

If the frequency of the dominant allele, B, is given by p and the frequency of the recessive allele, b, is given by q, and there are only two alleles, then the sum of the frequencies of the alleles is one:

p + q = 1

If we square both sides of the equation, then we get the frequencies of the different genotypes in the population:
p + q = 1
(p + q)^2 = 1^2
(p + q)^2 = 1
p^2 + 2pq + q^2 = 1
where p^2 is the frequency of dominant homozygotes, BB, 2pq is the frequency of heterozygotes, Bb, and q^2 is the frequency of recessive homozygotes, bb.

There are 84 + 16 = 100 moths in the population and 16 are white (bb), and so:

q^2 = 16/100 = 0.16
q = 0.4

p + q = 1
p = 1 - q
p = 1 - 0.4 = 0.6

p^2 = 0.6 x 0.6 = 0.36 (the frequency of the BB dominant homozygotes in the population).