If the radii of the circular …

Let us suppose that the Radius of the bigger end of the frustum (bucket) of cone = R = 20 cm
Suppose r denotes the Radius of the smaller end of the frustum (bucket) of the cone =  8 cm
Height = 16 cm
Now, Volume = {tex}\frac{1}{3} \pi{/tex}h[R² + r² + Rr]
= {tex}\frac{1}{3} \times \frac{22}{7} \times{/tex}16[20² + 8² + 20(8)]
= {tex}\frac{352}{21}{/tex}[400 + 64 + 160]
= {tex}\frac{352\times 624}{21}{/tex}
= {tex}\frac{219648}{21}{/tex}
= 10459.43 cm³
Now,
Slant height of the cone = l = {tex}\sqrt{(R -r)^2 + h^2}{/tex}
l = {tex}\sqrt{(20 -8)^2 +16^2}{/tex}
l = {tex}\sqrt{12^2 +16^2}{/tex}
l = {tex}\sqrt{144 + 256}{/tex}
l = {tex}\sqrt{400}{/tex}
l = 20 cm
Slant height is 20 cm
Now,
Surface area = {tex}\pi{/tex}[R² + r² + (R + r){tex}\times{/tex}l]
= {tex}\frac{22}{7}{/tex}[20² + 8² + (20 + 8){tex}\times{/tex}16]
= {tex}\frac{22}{7}{/tex}[400 + 64 + 448]
= {tex}\frac{22}{7} \times{/tex}912
= {tex}\frac{20064}{7}{/tex}
= 2866.29 cm² 

Thus, the surface area of the bucket is 28866.29