 Solution

Since $\beta$ is integral, $R[\beta ]$ is generated over $R$ by finitely many powers, hence is fractional. It is invertible by (c). Because it is a ring, $I^2=I$; multiplying by $I^{-1}$ gives $I=R$. Thus every $\beta \in {\mathcal{O}}_K$ lies in $R$, so $R={\mathcal{O}}_K$.