As per the inquiry,

The pail is as frustum of a cone.

We realize that,

Volume of frustum of a cone\[=\text{ }1/3\text{ }\pi h\left( r12\text{ }+\text{ }r22\text{ }+\text{ }r1r2 \right)\] , where, h = tallness, r1 and r2 are the radii(r1 > r2)

For pail,

Volume of pail = 28.490 L

1 L = 1000 cm3

Volume of pail = 28490 cm3

Sweep of top, r1 = 28 cm

Sweep of base, r2 = 21 cm

Let the tallness = h.

Subbing these qualities in the situation to discover the volume of container,

We have.

Volume of pail \[=\text{ }1/3\text{ }\pi h\left[ 282\text{ }+\text{ }212\text{ }+\text{ }28\left( 21 \right) \right]\]

\[28490\text{ }=\text{ }1/3\text{ }\times \text{ }22/7\text{ }\times \text{ }h\text{ }\left( 784\text{ }+\text{ }441\text{ }+\text{ }588 \right)\text{ }=\text{ }22/7\text{ }\times \text{ }h\text{ }\times \text{ }1813\]

\[\Rightarrow h\text{ }=\text{ }\left( 28490\times 21 \right)/\left( 22\times 1813 \right)\]

⇒ h = 15