As, per the question it is given
The internal radius of hollow sphere $=2cm$
The external radius of hollow sphere $=4cm$
As we know that,
Volume of the hollow sphere $4/3\pi \times \left( {{4}^{3}}-{{2}^{3}} \right)$ … (i)
It is given that,
The base radius of the cone $=4cm$
Assume the height of the cone be x cm
Volume of the cone $1/3\pi \times {{4}^{2}}\times h$ ….. (ii)
As the volume of the hollow sphere and cone are equal. We can equate equations (i) and (ii)
Then, we get
$4/3\pi \times \left( {{4}^{3}}-{{2}^{3}} \right)=1/3\pi \times {{4}^{2}}\times h$
$4\times \left( 64-8 \right)=16\times h$
$h=14$
Then,
Slant height of the cone (l) is given by
$l=\sqrt{\left( {{h}^{2}}+{{r}^{2}} \right)}$
$l=\sqrt{\left( {{14}^{2}}+{{4}^{2}} \right)}=\sqrt{212}$
$l=14.56cm$
Therefore, the height and slant height of the conical heap are $14cm$ and $14.56cm$ respectively.