jmc

Let $n$ be a positive integer. Then $n^2$ has exactly $3$ digits in base 9 if and only if $$9^2\le n^2<9^3.$$Taking square roots, we have $$3^2\le n<3^3.$$We are looking for $N$, the $\mathbf{largest}$ integer $n$ satisfying the above constraints. So, $$N=3^3-1=3\cdot 9-1=2\cdot 9+8.$$Written in base $9$, this is $\boxed{28}$ or $\boxed{28_9}$.