As per the question it is given that,
Height of the circular Cylinder $\left( {{h}_{1}} \right)=12cm$
Base radius of the circular Cylinder (r) $=5cm$
Height of the conical hole $=$ Height of the circular cylinder, i.e.,${{h}_{1}}={{h}_{2}}=12cm$
Then, Base radius of the conical hole $=$ Base radius of the circular Cylinder $=5cm$
Assume, L as the slant height of the conical hole.
Now, we all know that
$L=\sqrt{{{r}^{2}}+{{h}^{2}}}$
$L={{5}^{2}}+{{12}^{2}}$
$L=\sqrt{12+144}$
$L=13cm$
Then,
The total surface area of the remaining portion in the circular cylinder $\left( {{V}_{1}} \right)=\pi {{r}^{2}}+2\pi rh+\pi rl$
${{V}_{1}}=\pi {{\left( 5 \right)}^{2}}+2\pi \left( 5 \right)\left( 12 \right)+\pi \left( 5 \right)\left( 13 \right)$
${{V}_{1}}=210\pi c{{m}^{2}}$
Then, the volume of the remaining portion of the circular cylinder $=$ Volume of the cylinder $–$ Volume of the conical hole
$V=\pi {{r}^{2}}h-1/3\times 22/7\times {{r}^{2}}\times h$
$V=\pi {{\left( 5 \right)}^{2}}\left( 12 \right)-1/3\times 22/7\times {{5}^{2}}\times 12$
$V=200\pi c{{m}^{2}}$