Solution

We check the possible membership of each element in $A,B,C$.

Condition given is:

$(A\cap B)\subseteq C\subseteq (A\cup B)$

For one element:

If the element is in neither $A$ nor $B$, then it cannot be in $C$.
Number of choices $=1$

If the element is only in $A$, then it may or may not be in $C$.
Number of choices $=2$

If the element is only in $B$, then it may or may not be in $C$.
Number of choices $=2$

If the element is in both $A$ and $B$, then it must be in $C$.
Number of choices $=1$

Total choices for one element:

$1+2+2+1=6$

Since there are $n$ elements, total number of triples is:

$6^n$