Since the product of the ages is $72$, the ages must be one of the following combinations:
$$\begin{align*}
&2,2,18\\
&2,3,12\\
&2,4,9\\
&2,6,6\\
&3,3,8\\
&3,4,6\\
\end{align*}$$
The sums are $22,17,15,14,14$, and $13$ respectively. Since the sum and product of the ages didn’t give Jack enough information, the sum must have been $14$, the only possibility that admits more than one solution. The ages must therefore have been either $2,6$, and $6$ or $3,3$, and $8$. If they were $2,6$, and $6$, the two oldest would have been twins, and Bill (probably) wouldn’t have referred to his eldest child: it’s true that one of the six-year olds would technically have been his eldest child, but he’d probably have thought of them as being the same age. Jack inferred that the eldest child wasn’t a twin and concluded that Bill’s children were aged $3,3$, and $8$ years.
Added: Oops! As noted in the comments, $1$ is a possible age. That adds the sets $$\{1,1,72\},\{1,2,36\},\{1,3,24\},\{1,4,18\},\{1,6,12\},\{1,8,9\}$$ to the collection, with sums $74,39,28,23,19$, and $18$; fortunately, these add no further ambiguities.