Since every digit has equal likely chances of appearing at any of the three places in a 3 digit number

so out of 27 number

9 will have Unit digit 1, 9 will have unit digit 2 and 9 will have unit digit 3

So sum of the unit digits of all 27 numbers = 9(1+2+3) = 54

We write 4 at the unit digit place and 5 goes carry forward


so out of 27 number

9 will have Tens digit 1, 9 will have Tens digit 2 and 9 will have Tens digit 3

So sum of the Tens digits of all 27 numbers = 9(1+2+3) = 54

but 5 is carry forward so the sum becomes 54+5 = 59, We write 9 at the tens digit place and 5 goes carry forward
so far the sum looks like _ _ 9 4

This much calculation gives us the answer out of given options however we can do the same thing two more times to find exact sum which will be 5,994

Answer: E