Start by fixing the parents' genotypes. Cystic fibrosis needs two mutant alleles to show. Both parents look healthy, yet they have an affected child who must be homozygous for the mutation. That child got one mutant allele from each parent, which proves both parents are silent carriers.
Write the carrier as Aa, where A is the working allele and a is the faulty one. When two carriers have a child, simple Mendelian counting gives the genotypes AA, Aa, Aa, aa in equal quarters. Only the aa child shows the disease.
That single aa outcome out of four equally likely combinations gives a risk of one in four for any pregnancy. Genetics has no memory, so the previously affected daughter does not raise or lower the odds for the next baby; the chance stays the same each time.
Therefore the chance that the next child is affected is one quarter.
\[\boxed{\tfrac{1}{4}}\]