Let the radius of the drum be \(r\) cm. The radius of each cone-shaped iron piece is \[ \frac{32}{2} = 16 \text{ cm} \] According to the question: \[ \pi r^2 \times 12.8 = \frac{1}{3} \pi \times 16^2 \times 40 \times 3 \] Simplifying: \[ r^2 = \frac{16 \times 16 \times 40 \times 3}{3 \times 12.8} = \frac{800 \times 16}{12.8} = 800 \] So, \[ r = \sqrt{800} = 20\sqrt{2} \] Therefore, the diameter of the drum is \[ 2 \times 20\sqrt{2} = 40\sqrt{2} \text{ cm} \]