The answer is 0.41 or 41%.
According to the Hardy-Weinberg principle: p² + 2pq + q² = 1
where:
p² - frequency of homozygous dominant individuals (AA)
q² - frequency of homozygous recessive individuals (aa)
2pq - frequency of heterozygous individuals (Aa)
So, if there are 300 individuals in a population, and 58 of them are homozygous dominant individuals (AA), then:

Similarly, if there are 300 individuals in a population, and 120 of them are homozygous recessive individuals (aa), then:

Using the Hardy-Weinberg principle p² + 2pq + q² = 1, the frequency of heterozygous individuals (Aa) is 2pq = 1 - p² - q²:
2pq = 1 - 0.4 - 0.19
2pq = 0.41
Thus, the frequency of individuals with an Aa genotype in the population is 0.41 or 41%.