jmc

The last digit of a square must be either $1$, $4$, $5$, $6$, or $9$. Therefore, we only need to consider these squares. Only one square begins and ends with $1:121$. Similarly, one square begins and ends with $4:484$. No square begins and ends with $5$. One square begins and ends with $6:676$. No square begins and ends with $9$. Therefore, there are $\boxed{3}$ squares which are $3$-digit palindromes.