This can be solved with Punnett squares. The affected male has the

genotype, and the female is homozygous recessive. This means that she does not carry this, and therefore, will have the genotype XᵇXᵇ
While doing this Punnett Square, keep in mind that in order to have the male offspring affected, their genotype
has to be

Since there are no

, the probability of their male offspring inheriting this disease is 0%