A bucket of height 24 cm is in the form of frustum of a cone whose circular ends are of diameter 28 cm and 42 cm.

We have,

Height of the frustum, $h=24 \mathrm{~cm}$,

Radius of the open end, $R=\frac{42}{2}=21 \mathrm{~cm}$ and

Radius of the close end, $r=\frac{28}{2}=14 \mathrm{~cm}$

Now,

Volume of the bucket $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 24 \times\left(21^{2}+14^{2}+21 \times 14\right)$

$=\frac{176}{7} \times(441+196+294)$

$=\frac{176}{7} \times 931$

$=23408 \mathrm{~cm}^{3}$

$=23.408 \mathrm{~L} \quad\left(\mathrm{As}, 1000 \mathrm{~cm}^{3}=1 \mathrm{~L}\right)$

$\therefore$ The cost of the milk $=30 \times 23.408=₹ 702.24$

So, the cost of the milk which the bucket can hold is ₹702.24.