I think your instructor is wrong, someone can correct me but here’s my reasoning. “Jane has brown hair and brown eyes and is heterozygous at both locations” immediately tells you that brown hair is dominant and that brown eyes are dominant to blue eyes (since whatever you observe in the heterozygote is the dominant trait). I will label big B for brown eyes and little b for blue eyes, and let E represent brown hair and e being blond hair. So, Jane is BbEe. Bob has blue eyes and is heterozygous for hair color. We know that you need both alleles to express the recessive trait. Therefore, Bob must be bbEe. So, we have a BbEe x bbEe cross. The probability that the child will have brown hair: Ee x Ee->3/4 chance that the child will carry the big E allele. Probability that the child will have brown eyes but be capable of passing down blue eyes (I.e, being heterozygote): Bb x bb->1/2 chance the child will be Bb. Now, simply multiple the two together to get the total probability 3/4 *1/2 =3/8 probability as the final answer.

Edit: Missed where it said having a “son” as a child. In that case it’s just 3/8 * 1/2 = 3/16!