A bucket is in the form of a frustum of a cone with a capacity of $12308.8 {\mathrm{cm}}^3$ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the area of the metal sheet used in its making.

Given,
Frustum of a cone volume = $12308.8\mathrm{ }{\mathrm{cm}}^3$
Upper radius $\left(r_1\right)\mathrm{ is }20\mathrm{ cm}$.
Lower radius $\left(r_2\right)\mathrm{ is }12\mathrm{ cm}$.
We know that,
Frustum of cone volume$=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_1^2+{\mathrm{r}}_2^2+{\mathrm{r}}_1{\mathrm{r}}_2\right)$       (h=height)
$12308.8=\frac{\mathrm{\pi }}{3}\mathrm{h}(20\times 20+12\times 12+20\times 12)$
$12308.8\times 3=\mathrm{\pi h}(400+144+240)$
$12308.8\times 3=\frac{22}{7}\mathrm{h}\left(784\right)$
$h=\frac{12308.8\times 3\times 7}{22\times 784}$
h = 14.99

h = 15 cm (approx)
So, the slant height (l) of the bucket can be given as,
$l=\sqrt{h^2+{\left(r^1-r^2\right)}^2}$
$l=\sqrt{{15}^2+(20-12{)}^2}$
$l=\sqrt{225+(8{)}^2}$
$l=\sqrt{225+64}$
$l=\sqrt{289}$
$l=17\mathrm{ cm}$
Therefore, the surface area of the bucket can be given as,
$=\mathrm{\pi }\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)\mathrm{l}+{\mathrm{\pi r}}_2^2$
$=\mathrm{\pi }\left(20+12\right)17+\mathrm{\pi }{\left(12\right)}^2$
$=\mathrm{\pi }\left(32\right)17+\mathrm{\pi }{\left(12\right)}^2$
$$\begin{gathered} =\mathrm{\pi }\left(544+144\right) \\ =2162.28 {\mathrm{cm}}^2 \end{gathered}$$
Therefore, the area of the metal sheet is $\mathrm{2162.28}{ \mathrm{cm}}^2$.
Hence, option $\mathrm{2162.28}{ \mathrm{cm}}^2$ is correct.