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.