If a solid piece of iron in the form

(a) Given, dimensions of the cuboid = 49 cm x 33 cm x 24 cm

∴                   Volume of the cuboid = 49 x 33 x 24 = 38808 cm3

$[\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height $]$

Let the radius of the sphere is $r$, then

Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $\left[\because\right.$ voulme of the sphere $\left.=\frac{4}{3} \pi \times(\text { radius })^{3}\right]$

According to the question,

Volume of the sphere $=$ Volume of the cuboid

$\Rightarrow \quad \frac{4}{3} \pi r^{3}=38808$

$\Rightarrow \quad 4 \times \frac{22}{7} r^{3}=38808 \times 3$

$\Rightarrow \quad r^{3}=\frac{38808 \times 3 \times 7}{4 \times 22}=441 \times 21$

$\Rightarrow \quad r^{3}=21 \times 21 \times 21$

$\therefore \quad r=21 \mathrm{~cm}$

Hence, the radius of the sphere is $21 \mathrm{~cm}$.