Nice work, Daisy. The answer is E

I'll post my solution as well:
Begin by noting that, if the sum is 2, then there are two cases:
Case 1: The number has two 1’s and all remaining digits are 0
Case 2: The number has one 2 and all remaining digits are 0
If a number is less than 10^81, then that number has 81 digits or fewer. For this question, we’ll say that all numbers less than 10^81 have EXACTLY 81 digits, however some of those numbers have leading digits of zero (e.g., 0000…0001001 = 1001 and 0000…00200 = 200).
Case 1: In how many ways can we have an 81-digit number such that there are two 1’s and seventy-nine 0’s? Essentially, we have 81 places and we must select 2 of those places to hold the digit “1.” We can accomplish this selection in 81C2 (3240) ways.
So, there are 3240 numbers consisting of two 1’s

Case 2: In how many ways can we have an 81-digit number such that there is one 2 and the remaining digits are 0? We have 81 places and we must select 1 place to hold the 2 (the remaining places will be filled with 0’s). There are 81 places and we must select 1 place. We can accomplish this in 81 ways.
Adding the results from Cases 1 and 2, we get a total of 3321 numbers.