If you want to avoid misreading and misunderstanding, write "three-digit integers" 

    instead of "3 digit integers", as this standard form is traditionally accepted in Math.

Three-digit integers are from 100 to 999. In all, there are 999 - 99 = 900 such integer numbers.


The ones digits are from 0 to 9.  

    Each ones digit from 0 to 9 is repeated with the same frequency;

    the number of different digits is 10 and the frequency of each digit is 900/10 = 90.

    So, the sum of all digits in the ones position is 90 times (1 + 2 + 3 + . . . + 9) =  = 5*9 = 45.

    90 times 45 is  90*45 = 4050.  Thus the sum of all 900 ones digits is 4050.



The tens digits are from 0 to 9.  

    Each tens digit from 0 to 9 is repeated with the same frequency;

    the number of different digits is 10 and the frequency of each digit is 900/10 = 90.

    So, the sum of all digits in the tens position is 90 times (1 + 2 + 3 + . . . + 9) =  = 5*9 = 45.

    90 times 45 is  90*45 = 4050.  Thus the sum of all 900 tens digits is 4050.



The hundreds digits are from 1 to 9.  

    Each hundreds digit from 1 to 9 is repeated with the same frequency;

    the number of different digits is 9 (the zero is not included) and the frequency of each digit is 900/9 = 100.

    So, the sum of all digits in the hundreds position is 100 times (1 + 2 + 3 + . . . + 9) =  = 5*9 = 45.

    100 times 45 is  100*45 = 4500.  Thus the sum of all 900 hundreds digits is 4500.



Now the ANSWER to the problem's question is this sum  4050 + 4050 + 4500 = 12600.