Outer radius of the pipe = \(\cfrac{10}{2}\) cm = 5 cm Let the inner radius of the pipe be \(r\) cm ∴ Volume of the pipe = \(\pi (5^2 - r^2) \times 18\) cubic cm And volume of the solid iron sphere = \(\cfrac{4}{3} \pi \times 6^3\) cubic cm According to the question: \(\pi (25 - r^2) \times 18 = \cfrac{4}{3} \pi \times 216\) ⇒ \(25 - r^2 = 16\) ⇒ \(r^2 = 25 - 16 = 9\) ⇒ \(r = 3\) ∴ Inner radius of the pipe = 3 cm ∴ Thickness of the pipe = \(5 - 3 = 2\) cm