 Solution

Choose an ample divisor $A$. The Hodge index theorem for algebraic surfaces implies $D\cdot A>0$: if it were zero, $D^2=0$ would force the numerical class of $D$ to be zero. Consequently $-D$ cannot be effective, because every nonzero effective divisor has positive intersection with $A$, while $-D\cdot A<0$.

The Riemann–Roch theorem for algebraic surfaces and the preceding computation give $\chi (X,{\mathcal{O}}_X(D))=2+D^2/2=2$. By Serre duality, $h^2(X,D)=h^0(X,-D)=0$. Thus