An autosomal recessive trait has a frequency of 1:1000. What will be the carrier frequency if the population is in Hardy-Weinberg equilibrium?

The correct option is B 0.058
If the population is in Hardy-Weinberg equilibrium, then AA is $p^2$, Aa is 2pq, and aa is $q^2$
Since the trait is recessive and one in thousand individuals expressed the trait
q = $\sqrt{q^2}=\sqrt{\frac{1}{1000}}$
q = $\sqrt{0.001}$
q = 0.031

We know that p + q = 1
p + 0.03 = 1
p = 1- 0.03
p = 0.97

Frequency of heterozygotes(carriers) are 2pq
2pq = $2\times 0.97\times 0.03$
2pq = 0.058

So the carrier frequency in the population is 0.058.