 Solution

Riemann-Roch theorem is $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$. Putting $D=0$ and using $\ell (0)=1$, $\ell (K)=g$ gives $\mathrm{deg}K=2g-2$. If an effective degree-two divisor has $\ell (D)=2$, its pencil defines a degree-two map to ${\mathbb{P}}^1$, so the curve is hyperelliptic. For $g=2$, Riemann-Roch applied to $K$ supplies such a degree-two pencil. For nonhyperelliptic genus 3, the canonical system has dimension 2, separates points and tangents, and hence embeds $C$ as a plane quartic. On a genus-3 hyperelliptic curve, $\ell (2Q)=2$ exactly at the $2g+2=8$ ramification points.