(a) 1 cm
Let the internal radius of the tube be r cm.
External radius = \(\frac82\) cm = 4 cm,
Height = 20 cm
∴ Volume of iron used in the tube = π(42 - r2) x 20 cm
Given, π x (16 - r2) x 20 = 440
⇒ \(\frac{22}{7}\times20\times(16-r^2) = 440\)
⇒ 16 – r2 = 7 ⇒ r2 = 16 – 7 = 9 ⇒ r = 3
∴ Thickness of tube = 4 cm – 3 cm = 1 cm.