Look at the circumference of the unit disk. Each of the disks must be capable of covering up at least $\frac{1}{3}$ of the circumference, which means it must be able to cover a chord of length $\sqrt{3}$. Thus, $\frac{\sqrt{3}}{2}$ is a lower bound for $r$. This bound is achievable: place the three centers of the disks symmetrically at a distance of $\frac{1}{2}$ from the center of the unit disk.