A bucket is in the form of a frustum of a cone

We have,

Radius of upper end, $R=28 \mathrm{~cm}$ and

Radius of lower end, $r=21 \mathrm{~cm}$

Let the height of the bucket be $h$.

Now,

Volume of water the bucket can hold $=28.49 \mathrm{~L}$

$\Rightarrow$ Volume of bucket $=28490 \mathrm{~cm}^{3} \quad\left(\mathrm{As}, 1 \mathrm{~L}=1000 \mathrm{~cm}^{3}\right)$

$\Rightarrow \frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)=28490$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times\left(28^{2}+21^{2}+28 \times 21\right)=28490$

$\Rightarrow \frac{22 h}{21} \times 49 \times(16+9+12)=28490$

$\Rightarrow \frac{22 h}{3} \times 7 \times 37=28490$

$\Rightarrow h=\frac{28490 \times 3}{22 \times 7 \times 37}$

$\therefore h=15 \mathrm{~cm}$

So, the height of the bucket is 15 cm.