The answer is; 48%
This is how to calculate;
Let's assign the dominant allele letter ‘T’ and the recessive letter ‘t’
Using the hardy weignburg equation; p2 + 2pq + q2 = 1
P2 + 2pq = 0.64
q2 = 1 – 0.64 = 0.36
q = 0.6
Remember that; p + q = 1
P = 1 – 0.6 = 0.4
Now to calculate population with herozygour trait = 2pq
2 * 0.4 * 0.6 = 0.48
This means 48% of the population is heterozygous for the trait that enable tasting of phenylthiocarbamide