jmc

$$\begin{array}{rl}(5^2{)}^7\times (2^3{)}^3& =5^{14}\times 2^9\\ & =5^5\times 10^9\end{array}$$ Since $5^5$ doesn't end in a $0$, we can conclude that the product ends in $9$ zeroes $\to \boxed{9}$.