In a family with 3 kids, the number $X$ of sons follows a binomial distribution with parameters $n=3$ and $p=0.5$ (good approximation :-) So the probability that this family has two sons and a daughter is ${\rm P}(X=2)=\binom32\frac{1}{2^3}=\frac38$.
Now for $2000$ families, the number $Y$ of families having two sons and a daughter follows a binomial distribution with parameters $n=2000$ and $p=\frac38$. The expectation of this random variable is ${\rm E}(Y)=np=2000\times\frac38 = 750$.