The statement is false.
Let the original radius be \(r\), now it becomes \(2r\), and let the original volume be \(v\), and the new volume be \(V\). \(\therefore\) \(\cfrac{V}{v} = \cfrac{\cfrac{4}{3}\pi (2r)^3}{\cfrac{4}{3}\pi (r)^3} = \cfrac{8r^3}{r^3} = \cfrac{8}{1}\) \(\therefore\) \(V = 8 \times v\)