= Solution
<Riemann-Roch theorem> is $\ell(D)-\ell(K-D)=\deg D+1-g$. Putting $D=0$ and using $\ell(0)=1$, $\ell(K)=g$ gives $\deg K=2g-2$. 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.
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