smc

A perfect square will be $(8k+r{)}^2=64k^2+16kr+r^2\equiv r^2(\mathrm{mod}16)$ where $r=0,1,...,7$. Notice that $r^2\equiv 1,4,9,0(\mathrm{mod}16)$. Now $ab3c$ in base 8 is $a{8}^3+b{8}^2+3(8)+c\equiv 8+c(\mathrm{mod}16)$. It being a perfect square means $8+c\equiv 1,4,9,0(\mathrm{mod}16)$. That means that c can only be 1 so the answer is 1 = $\boxed{1}$.