= Solution
First form the <singleton set> $\{b\}=F_1(b,b)$ and then use $F_2$ to take a <set union>:
$$
U(a,b)=F_2\bigl(F_1(a,F_1(b,b)),a\bigr)=\bigcup\{a,\{b\}\}=a\cup\{b\}.
$$
The unused second argument of $F_2$ may be any term. Since $x\cap c=x\setminus(x\setminus c)$, a term using only the prescribed operation symbols is
$$
\boxed{G(a,b,c)=F_3\bigl(U(a,b),F_3(U(a,b),c)\bigr).}
$$
After substituting the displayed term for both occurrences of $U$, this is literally a term in $F_1,F_2,F_3$, and its value is $(a\cup\{b\})\cap c$.
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