Answer:
Since the condition is sex-linked, it must be related to the X chromosome.
Let [tex]X^{A}[/tex] represent the dominant allele and [tex]X^{a}[/tex] the recessive one, the daugther must have a [tex]X^{a}[/tex] chromosome from the father (the only one he could pass) and a [tex]X^{A}[/tex] from the mother, otherwise she wouldn't be female ([tex]X^{a}Y[/tex]) or would have the condition ([tex]X^{a}X^{a}[/tex]).
If the man does not have the condition, he has the dominant allele on the X chromosome, represented as [tex]X^{A}Y[/tex]. You can use Punett squares to represent their cross.
\begin{center}\begin{tabular}{ |c|c|c|c| }\ & X^{a} & X^{A} \\ \ X^{A} & X^{A}X^{a} & X^{A}X^{A} \\ \ Y & X^{a}Y & X^{A}Y \\ \end{tabular}\end{center}
None of the possible daugthers could have the condition, but there is a 50% chance of their son having it (1/2).
In total, since there is a chance of 1/2 for their son to be affected, the probability of having four sons with the condition is 1/2⁴, or 6.25%.
