Answer: C
Radius of the hemispherical bowl = \(\frac{1.4}{2}\) dm = 0.7 dm \(\therefore\) Outer surface area of the bowl = \(2\pi (0.7)^2\) square dm \(= 2 \times \frac{22}{\cancel{7}} \times \frac{\cancel{49}7}{100}\) square dm \(= 3.08\) square decimeters.
Radius of the hemispherical bowl = \(\frac{1.4}{2}\) dm = 0.7 dm \(\therefore\) Outer surface area of the bowl = \(2\pi (0.7)^2\) square dm \(= 2 \times \frac{22}{\cancel{7}} \times \frac{\cancel{49}7}{100}\) square dm \(= 3.08\) square decimeters.