Update: Due to the down-votes I took the time to carefully explain the method here.
Note: If you remember your times tables you can actually do the work written out below in your head using your fingers to keep track of where you are at (yes, you don't have 16 fingers but it's still doable).
Moreover, if you observe that the OP's answers b) and c) both have two leading zeroes, you can stop after the third calculation with the $\text{*}$ indicator below, saving you one last calculation.
Working in $\Bbb N \to \{0, 1,2,3, \dots\}$.
Any integer $a \ge 1$ has a $\text{base-}10$ representation,
$\tag 1 a = \displaystyle{\sum_{k=0}^n a_k 10^k} \text{ with } a_k \in \{0, 1,2,3,4,5,6,7,8,9\} \text{ and } a_n \gt 0$
We define $\rho: \Bbb N^{\gt 0} \to \{1,2,3,4,5,6,7,8,9\}$ as follows:
$\quad \rho(n) = \text{the smallest } k \text{ such that } a_k \ne 0 \text{ in (1)}$
For the OP numbers we have
$\quad \rho(20 \; 922 \; 789 \; 888 \; 000) = 8$
$\quad \rho(18 \; 122 \; 471 \; 235 \; 500) = 5$
$\quad \rho(17 \; 223 \; 258 \; 843 \; 600) = 6$
Proposition: $\rho(ab) = \rho\big(\rho(a)\rho(b)\big)$.
Calculations
$\rho(1!) = 1$
$\rho(2!) = 2$
$\rho(3!) = 6$
$\rho(4!) = 4$
$\rho(5!) = 2$ *
$\rho(6!) = 2$
$\rho(7!) = 4$
$\rho(8!) = 2$
$\rho(9!) = 8$
$\rho(10!) = 8$ *
$\rho(11!) = 8$
$\rho(12!) = 6$
$\rho(13!) = 8$
$\rho(14!) = 2$
$\rho(15!) = 3$ *
$\rho(16!) = 8$
So the answer is a).