The number is too damm high

Let $A = (10 + \sqrt{90})^k$ where $k$ is the exponent given in the problem. The only important properties of $k$ are that it is big and it is even. We may write $A = \sum_{i=0}^k \binom{k}{i} \sqrt{90}^i 10^{k-i}$ . Let's also define $B = (10 - \sqrt{90})^k = \sum_{i=0}^k \binom{k}{i} (-1)^i \sqrt{90}^i 10^{k-i}$ . Notice that $A+B = 2\sum_{i=0}^{\frac{k}{2}} \binom{k}{2i} \sqrt{90}^{2i} 10^{k-2i} = 2 \sum_{i=0}^{\frac{k}{2}} \binom{k}{2i} 90^i 10^{k-2i}$ . What is important about this number is that it is an integer. And thus $\lfloor A \rfloor = \lfloor A + B - B \rfloor = A + B + \lfloor - B \rfloor$ . Because $B$ is a positive number less than 1 we already know that $\lfloor -B \rfloor = -1$ so the previous expression is just $A + B - 1$ .

To find the second digit just compute $A + B - 1 \equiv -1 \equiv 99 \mod 100$ . The answer is 9.