The correct option is B 96
If frequency q of the autosomal lethal is 0.4, then p = 1-q = 0.6.
Following the Hardy-Weinberg equation p^2 + 2pq + q^2 = 1,
p^2 = 0.36, 2pq = 0.48, q^2 = 0.16.
2pq is the frequency of heterozyogte carriers, so 0.48 x 200 = 96 individuals.