The problem is as follows:
How many zeroes do we write when we write all the integers from 1 to 243 in base 3?
The given solution starts as follows:
The 1-digit numbers don't have any zeroes.
The 2-digit numbers use 2 zeroes: 10 and 20.
There are $3^2 = 9$ three-digit numbers starting with 1 and 9 starting with 2. For each leading digit, a zero appears in each digit in 9/3 = 3 of the numbers, so each has a total of 3+3 = 6 zeroes. Thus, the 3-digit numbers contain 2*6 = 12 zeroes.
I do not understand how they arrived with the fact that
"for each leading digit a zero appears in 9/3 = 3 of the numbers"
and why they are multiplying the result by 2 to get 12 at the end.