In a Hardy Weinberg equilibrium :-->
p^2 + 2pq + q^2 = 1

Here , p = black , q = grey .
p^2 = 9 So p = 3 and if p = 3 then q = 7 (because p+q = 1) frequency ...

So , Heterozygous = 2pq
= 2×3×7 = 42

So , frequency of heterozygous = 0.42 ...ans...

$$Hope it's help... $$

Given Information:
- In a diploid population of Drosophila, 9% of the population is black in color.
- Black color is recessive to grey color.

Hardy-Weinberg Equilibrium:
The Hardy-Weinberg equilibrium describes the relationship between the frequencies of alleles in a population and the frequencies of genotypes. According to this equilibrium, the allele and genotype frequencies in a population will remain constant from generation to generation if certain assumptions are met. These assumptions include no mutation, no migration, no selection, random mating, and a large population size.

Calculating Allele Frequencies:
Let's assume that the black color allele is represented by "b" and the grey color allele is represented by "B". Since black color is recessive, the black flies must be homozygous for the black allele (bb), while the grey flies can be either homozygous for the grey allele (BB) or heterozygous (Bb).

Let p = frequency of the B allele (grey color allele)
Let q = frequency of the b allele (black color allele)

According to the Hardy-Weinberg equilibrium, the sum of allele frequencies must be equal to 1. Therefore, p + q = 1.

Since black color is recessive, the frequency of the black flies (bb) can be calculated using the equation: q^2 = 0.09, where q^2 represents the frequency of the bb genotype.

Solving for q, we get:
q^2 = 0.09
q = √(0.09)
q = 0.3

Since p + q = 1, we can calculate the frequency of the grey allele (p):
p = 1 - q
p = 1 - 0.3
p = 0.7

Calculating the Frequency of Heterozygous Flies:
The frequency of heterozygous flies (Bb genotype) can be calculated using the equation: 2pq, where p and q represent the frequencies of the alleles.

Substituting the values, we get:
2pq = 2(0.7)(0.3)
2pq = 0.42

Therefore, the frequency of heterozygous flies in the population is 0.42.

Answer:
The correct option is A. 0.42.