A spherical shell of lead, whose external and internal diameters are 24 cm and 18 cm,

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell $=\frac{4}{3} \pi\left(12^{3}-9^{3}\right)=\frac{4}{3} \pi(1728-729)=\frac{4}{3} \pi \times(999)=4 \pi \times(333) \mathrm{cm}^{3}$

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder $=\pi r^{2} \mathrm{~h}=37 \pi r^{2} \mathrm{~cm}^{3}$

Volume of the shell = Volume of cylinder 

$O r, 4 \pi \times(333)=37 \pi r^{2}$

$\Rightarrow r^{2}=\frac{4 \times 333}{37}=4 \times 9$

$\Rightarrow r=\sqrt{4 \times 9}=\sqrt{36}=6 \mathrm{~cm}$

So, diameter of the base of the cylinder = 2r = 12 cm.