We know an $n$-Sudoku puzzle is with $n \times n$ subgrids consisting of $n \times n$ cells; you will fill it with numbers from $1$ to $n^2$.
Candidate solution have size polynomial in $n$, and can be checked in a time polynomial in $n$. The number of cells filled in each solution is at most $n^2 \times n^2 = n^4$. The length of each number is $\log(n^2) = 2\log n$. So the length of each solution is $O(n^4\log n)$, which is $O(\leq n^5)$, since $\log n \leq n$. (The reason is that $2^n \geq 1 + n$ by the binomial theorem, so $2^n \geq n$, and we then take $\log_2$ of both sides.)
My only question is why the number of each number is $\log(n^2)$.
We need at most $\lceil \log_2 N\rceil+1$ binary digits in order to present a natural number $N$ in a binary system.