A metallic spherical shell of internal and external diameters 4 cm and 8 cm,

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.

Volume of the spherical shell= $\frac{4}{3} \pi\left(R^{3}-r^{3}\right)$

$=\frac{4}{3} \pi\left[(4)^{3}-(2)^{3}\right] \mathrm{cm}^{3} \quad\left[\right.$ Since $\left.\mathrm{r} \Rightarrow \frac{4}{2}=2 \mathrm{~cm}, \mathrm{R}=\frac{8}{2}=4 \mathrm{~cm}\right]$

$=\left(\frac{4}{3} \pi \times 56\right) \mathrm{cm}^{3}$

Radius of the cone $=\frac{8}{2}=4 \mathrm{~cm}$

Volume of the cone $=\frac{1}{3} \pi r^{2} h$

$=\left(\frac{1}{3} \pi \times 4 \times 4 \times h\right) \mathrm{cm}^{3}$

Therefore,

$\frac{1}{3} \pi \times 4 \times 4 \times h=\frac{4}{3} \pi \times 56$

$\Rightarrow \frac{16}{3} \times h=\frac{4}{3} \pi \times 56$

$\Rightarrow h=\frac{4 \times 56 \times 3}{3 \times 16}$

$\Rightarrow h=14 \mathrm{~cm}$