2 Floors are harder than 9 Floors?

The sum of odd numbers (starting from 1) up to the $n^\text{th}$ term is $\frac{\big(1+(2n-1)\big)\times n}{2} = \frac{2n^2}{2}=n^2.$ Now, let $k$ be the number of floors that the building has. Then we have $\begin{aligned} 9^2 < &k^2-(k-2)^2\\ 81 < &4k-4\\ 85 < &4k\\ 21.25 < &k \le 25 && (\text{because of height restrictions})\\ \Rightarrow &k=22, {\color{#D61F06}{23}}, 24, 25. && (\text{because it's prime})\\ \end{aligned}$ Therefore, the building has 23 floors. $_\square$

Knowing that the answer was unique, we only need to maximize the number of floors. The greatest prime number less than 25 is $\boxed{23}$

Let $p$ be the number of floors. All we are given is that $p\le25$ and "The combined calorie burning in the top 2 floors is greater than the total calorie burning in the first 9 floors", so we know this must be enough information to solve the problem. When you think about it, you'll realize that the "the total calorie burning in the first 9 floors" is constant while "The combined calorie burning in the top 2 floors" increases as $p$ increases. This means that if this statements holds for $p$ , then must also hold for any prime greater than $p$ ! (<--- an exclamation point, not a factorial). Thus, assuming this question has a well-defined answer, it must be the largest prime $p\le25$ and so the answer is 23.