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

$V=\pi {{r}^{2}}h-1/3\times 22/7\times {{r}^{2}}\times h$

$V=\pi {{\left( 5 \right)}^{2}}\left( 12 \right)-1/3\times 22/7\times {{5}^{2}}\times 12$

$V=200\pi c{{m}^{2}}$