= Solution
Assume a single disease <allele> $a$, complete <penetrance> of $aa$, no disease in $Aa$ or $AA$, and <Hardy-Weinberg equilibrium> in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.
For a child with <inbreeding coefficient> $F$, the two inherited copies are <identical by descent> with <probability> $F$. In that component there is just one ancestral <allele> draw, and it is $a$ with <probability> $q$. In the remaining component there are two independent ancestral draws, with <probability> $q^2$ of $aa$. Thus the <inbreeding genotype frequencies> give
$$
P(aa)=Fq+(1-F)q^2=q^2+Fq(1-q).
$$
Using the <first-cousin inbreeding coefficient> $F=1/16$, the disease-risk ratio is
$$
\boxed{\frac{P(aa)}{q^2}=1+\frac{F(1-q)}q=1+\frac{999}{16}=63.4375\simeq63.4.}
$$
The absolute risk in this model is $6.34375\times10^{-5}$. This calculation averages over the parental <genotypes> in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known <genetic carriers>, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease <pedigree>.
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