jmc

Let $\theta =\angle ABC$. The base of the cylinder is a circle with circumference 6, so the radius of the base is $6/(2\pi )=3/\pi$. The height of the cylinder is the altitude of the rhombus, which is $6\sin \theta$. Thus the volume of the cylinder is $$6=\pi {\left(\frac{3}{\pi }\right)}^2\left(6\sin \theta \right)=\frac{54}{\pi }\sin \theta ,$$ so $\sin \theta =\boxed{\frac{\pi }{9}}$.