Solution:

$I$ always lies inside the convex hull of $ABC$, which in turn always lies in the circumcircle of $ABC$, so $IO < R$. On the other hand, if we first draw the circle $\Omega$ of radius $R$ about $O$ and then pick $A$, $B$, and $C$ very close together on it, we can force the convex hull of $ABC$ to lie outside the circle of radius $R - \epsilon$ about $O$ for any $\epsilon$. Thus the answer is $R$.