I tried to solve like this.
Imagine(please imagine) we have T(4) and I have computed values of T(1) and T(0) only.
So we need to compute T(4) T(3) and T(2). And when it reach T(1)., T(4) T(3) and T(2) will be in stack.
So the no of values present in the stack can be written as 4-2+1=3.(I am adding 1 in the eq because I am including T(2) too).
Now moving back to the question, suppose we have T(13). and in the question they have mentioned that they have computed values of T(1) and T(0). So using the same argument we can say that we have
13-2+1=12 records in the stack which is the max. So 13 should be the answer.