The short way of doing this is recognizing that the numbers are occurring in sets of 3, which means at least one 3 is factorable in the sum.

So the answer choice must be at least a multiple of 3. Choice E fits that puzzle.

the long way of doing this averaging out evenly set numbers.

111+112+113+121+122+123+...

112*3 + 122*3 + ...132*3

you know there are 27 numbers, you can categorize them in groups of 3 and you will have 9 sets.

so 3*(112 + 122 + 132 + 212 + 222 + 232...)

these 9 sets of numbers can be further categorized into 3 sets of 3.

3*(122*3 + 222*3 + 332*3)

3*3(122 + 222 + 332)

the final set is also an evenly spaced set, so you can further simplify it into:

3*3*(222*3) = 3*3*3(222) = 5994

so ultimately the sum has to be a multiple of 27 which 5994 is.

(E)

Here's another way:

Each column is going to add up to 54. We can figure this out by seeing that since there are 27 numbers in the column, we'll have 9 of each number (27/3), so (9x1)+(9x2)+(9x3)=54.

So, now just add. The units' digit column adds to 54, so the units' digit is 4 and we carry the 5.

The tens' column also adds to 54, but we carried the 5, so it's 59. The tens' digit is 9, and we carry the 5.

The hundreds' column also adds to 54, but we carried the 5, so it's 59.

And we have:

5994

EDIT -- Not that this saves much time in this case, but we can stop at the tens' digit since there's only one answer choice that ends in 94. Could be helpful on a similar problem.