Proof. It is easy to see that for each point $x\in B^{sing}$ one has $i_{loc}(x)={\frac{1}{12}}$. Then from Gauss-Bonnet theorem one deduces that $\chi(B)=2-2g$, where $g$ is the genus of the Riemann surface $B$. Then we have

$${\frac{|B^{sing}|}{12}}=2-2g\,\,.$$

Since LHS is non-negative we conclude that either $g=1$ or $g=0$. In the first case $|B^{sing}|=0$ and we have a ${\bf Z}$-affine structure on a torus. In the second case we have $|B^{sing}|=24$ and $g=0$. $\blacksquare$