Answer: Outer surface area = 452.16 cm²; mass ≈ 3383 g (π = 3.14)
Outer radius R = 6 cm; thickness 0.01 m = 1 cm, so inner radius r = 5 cm.
\[\text{Outer surface}=4\pi R^2=4\times3.14\times36=452.16\ \text{cm}^2\]
\[V_{\text{metal}}=\tfrac43\pi(R^3-r^3)=\tfrac43\times3.14\times(216-125)\]
\[=\tfrac43\times3.14\times91\approx380.99\ \text{cm}^3\]
\[\text{Mass}=380.99\times8.88\approx3383.16\ \text{g}\]
(About 3.38 kg. With \(\pi=\tfrac{22}{7}\) the mass is about 3386.24 g; the textbook key gives 3385.94 gm.)