Let the length of the belt be $b$.
^2-(2r)^2} =2\sqrt{3}r ,\\\ \phi&=\arccos\tfrac{2r}{4r}=\tfrac\pi3 ,\\\ b&=2\,((\pi-\phi)\cdot3r+d+\phi\,r) \\\ &=2(2\pi r+2\sqrt{3}r+\tfrac\pi3 r) \\\ &=(\tfrac{14}3 \pi +4\sqrt{3})r ,\\\ r&=\frac{b}{\tfrac{14}3 \pi +4\sqrt{3}} . \end{align}