By considering modulo $5$ and modulo $4$, only $00$ and $44$ are possible last digits. Let the square be $n^2$. So $n \ge 100$ and $n \le 316$. The first combination occurs for $n^2$ iff $n=10k$ is a multiple of $10$. So we need to count three digit squares $k^2$. That's $k \ge 10$ and $k \le 31$ so $31-9=22$ solutions. The second combination occurs iff $n \equiv \pm 12 \mod 100$. So we have the solutions $n=112, n=212, n=312$ and $n=188, n=288$. That's $5$ solutions, adding up to a total of $27$ solutions.

Isn't n congruent to +12 or - 12 mod 50 in the second case? That would give us 4 more solutions :n=138, n=162,n=238,n=262.