If the base $8$ representation of a perfect square is $ab3c$, where $a\ne 0$, then $c$ equals 
$\text{(A)} 0\qquad  \text{(B)}1 \qquad  \text{(C)} 3\qquad  \text{(D)} 4\qquad  \text{(E)} \text{not uniquely determined}$
A perfect square will be $(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}$ where $r=0,1,...,7$.
Notice that $r^2\equiv 1,4,9,0 \pmod{16}$.
Now $ab3c$ in base 8 is $a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}$. It being a perfect square means $8+c\equiv 1,4,9,0 \pmod{16}$. That means that c can only be 1 so the answer is 1 = $\boxed{\textbf{(B)}.}$.