= Solution
Use <identity-by-descent sharing of inbred full siblings>, with $J$ the maximum number of copies that can be paired across the children as <identical by descent>. If two copies within one child have the same ancestral origin, they still count as two copies when both can be paired.
With probability $3/4$ the cousin parents share no ancestral copy. Conditional on this event, their children have ordinary <Mendelian IBD sharing of full siblings>, namely $(1/4,1/2,1/4)$. With probability $1/4$, the parental copies can be labelled $(A,B)$ and $(A,C)$, with $A,B,C$ distinct ancestral origins. Each child independently receives one of $AA,AB,AC,BC$, with probability $1/4$ each. Among the sixteen ordered child pairs, $AA$ paired with $BC$ in either order gives $J=0$; four equal genotype-origin pairs give $J=2$; the remaining ten give $J=1$. The mixture is
$$
\boxed{\begin{aligned}P(J=0)&=\frac34\frac14+\frac14\frac18=\frac7{32},\\P(J=1)&=\frac34\frac12+\frac14\frac58=\frac{17}{32},\\P(J=2)&=\frac34\frac14+\frac14\frac14=\frac14.\end{aligned}}
$$
These are inbred children, so the three sharing probabilities do not retain all within-child identities. The full <Jacquard identity coefficients> are $(1,0,2,1,2,1,15,30,12)/64$. In particular, their <kinship coefficient> is $9/32$; applying the outbred formula $(P(J=1)+2P(J=2))/4$ here would be incorrect.
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