A hollow sphere of internal and external diameters 4 cm and 8 cm

External radius $r_{\mathrm{i}}=\frac{8}{2}=4 \mathrm{~cm}$

Internal radius $r_{2}=\frac{4}{2}=2 \mathrm{~cm}$

The volume of hollow sphere

$V=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)$

$=\frac{4}{3} \pi\left(4^{3}-2^{3}\right)$

Let h be the height of cone.

Clearly,

The volume of recasted cone = volume of hollow sphere

$\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi\left(4^{3}-2^{3}\right)$

$\Rightarrow 4^{2} \mathrm{~h}=4\left(4^{3}-2^{3}\right)$

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

Hence, the height of cone = 14 cm

Hence, the correct answer is choice (b).