Solution: The volume of the sphere is $ \frac{4}{3}\pi (2x)^3 = \frac{32}{3}\pi x^3 $. The volume of the hemisphere is $ \frac{2}{3}\pi (3x)^3 = 18\pi x^3 $. The ratio is $ \frac{32/3}{18} = \frac{32}{54} = \frac{16}{27} $. The ratio of their volumes is $\boxed{\dfrac{16}{27}}$.

["Understanding the Volume Ratio Between a Sphere and a Hemisphere: A Clear Explanation", "When exploring geometric shapes, understanding volume relationships can deepen your comprehension of spatial mathematics. In this article, we explore the volume ratio between a sphere and a hemisphere with a practical example to clarify common calculation pitfalls.", "Step 1: Calculate the Volume of the Sphere\nThe formula for the volume of a sphere is:\n[\nV_{\ ext{sphere}} = \frac{4}{3}\pi r^3\n]\nGiven the sphere has radius ( 2x ), substitute ( r = 2x ):\n[\nV_{\ ext{sphere}} = \frac{4}{3}\pi (2x)^3 = \frac{4}{3}\pi (8x^3) = \frac{32}{3}\pi x^3\n]", "Step 2: Calculate the Volume of the Hemisphere\nThe volume of a full sphere is ( \frac{4}{3}\pi r^3 ), so a hemisphere (half of a sphere) is:\n[\nV_{\ ext{hemisphere}} = \frac{1}{2} \ imes \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3\n]\nWith radius ( 3x ), substitute ( r = 3x ):\n[\nV_{\ ext{hemisphere}} = \frac{2}{3}\pi (3x)^3 = \frac{2}{3}\pi (27x^3) = 18\pi x^3\n]", "Step 3: Determine the Volume Ratio\nNow, compute the ratio of the sphere’s volume to the hemisphere’s volume:\n[\n\ ext{Ratio} = \frac{V_{\ ext{sphere}}}{V_{\ ext{hemisphere}}} = \frac{\frac{32}{3}\pi x^3}{18\pi x^3}\n]\nCancel ( \pi x^3 ) from numerator and denominator:\n[\n= \frac{32}{3} \div 18 = \frac{32}{3} \ imes \frac{1}{18} = \frac{32}{54}\n]\nSimplify the fraction:\n[\n\frac{32}{54} = \frac{16}{27}\n]", "Conclusion:\nThe volume ratio of the sphere (radius ( 2x )) to the hemisphere (radius ( 3x )) is:\n[\n\boxed{\dfrac{16}{27}}\n]\nThis result demonstrates how scaling radius affects three-dimensional volume, reinforcing key principles in geometry.", "Understanding these relationships not only helps solve textbook problems but also improves spatial reasoning—useful in engineering, architecture, and design fields. When comparing volumes of curved bodies like spheres and hemispheres, exact calculations ensure accuracy in planning and material estimation.", "---", "By breaking down each formula application and simplification step, readers gain both clarity and confidence in solving volume-related geometry problems."]