Yes, that number does seem too large, doesn't it?
It appears you used a formula to calculate the sum of the first 20 terms of the progression, S(20). This isn't relevant to your situation, where you just seek the 20th term.
The first term is 110. To get the next you would multiply this by 1.06. So the second term would be 110*(1.06)^(2-1).
I wrote it that way to demonstrate an example of the general formula for the nth term, which is 110*(1.06)^(n-1).
So for your situation, the 20th term would be 110*(1.06)^(20-1) = 332.8, a much more reasonable number for the situation.
Hope this helps, and if you like here is the relevant formula [see below] used when working with geometric progression problems like this one.
a(n) = a*r^(n-1), where 'a' is the first term [110 itt], and 'r' is the common ratio [1.06 itt]