Find zeros

$\begin{aligned} &= 2^{13} \times 3^4 \times 5^{18} \times 251^3 \\ &= 3^{4} \times 5^{5} \times 251^{3} \times (10^{13}) \implies \boxed{13} \end{aligned}$

Taking $2008^3 = 2^{9}251^{3}$ and $25^5 = 5^{10}$ we obtain $2008^3 \cdot 25^5 = (2 \cdot 5)^{9} 5^1 251^3 = (5^1 251^3) \cdot 10^9$ . Now the final product can compute to:

$(5^1 251^3) \cdot 10^9 \times (15^4) \cdot 10^4 = (3^4 5^5 251^3) \cdot 10^{13} \Rightarrow \boxed{13}$ trailing zeros.