Each of the five options given is a member of the sequence, so each is a multiple of 9 (because they are multiples of 3 x 3).
So if the number is going to be a multiple of 81, we require the third factor of the number (the factor consisting of several 2s followed by a single 3), to also be a multiple of 9.
A number is divisible by 9 if and only if the sum of its digits is a multiple of 9.
For each number in the sequence, the number of 2s in its third factor equals the number of 0s in that number.
The options given have 4, 6, 8, 10 or 12 zeros, corresponding with their third factors having digit sums of 11, 15, 19, 23 and 27 respectively.
Of these, only 27 is a multiple of 9 so the correct answer is 20,000,000,000,007.
This problem is taken from the UKMT Mathematical Challenges.