A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed? 
$\mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\%$
The volume of the original box is $15\cdot10\cdot8=1200.$
The volume of each cube that is removed is $3\cdot3\cdot3=27.$
Since there are $8$ corners on the box, $8$ cubes are removed. 
So the total volume removed is $8\cdot27=216$. 
Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}.$