In a population with two alleles B and b with frequency 0.6 and 0.4...

According to the Hardy-Weinberg principle, the frequency of alleles in a population is represented by $p$ and $q$.

Given the frequency of allele B is $p = 0.6$ and the frequency of allele b is $q = 0.4$.

The frequency of heterozygous individuals is given by $2pq$.

$2pq = 2 \times 0.6 \times 0.4 = 0.48$

The expected number of heterozygous individuals in a sample of $250$ is $2pq \times 250$.

$0.48 \times 250 = 120$

Answer: $120$