smc

The volume of the original box is $15\cdot 10\cdot 8=1200.$ The volume of each cube that is removed is $3\cdot 3\cdot 3=27.$ Since there are $8$ corners on the box, $8$ cubes are removed. So the total volume removed is $8\cdot 27=216$. Therefore, the desired percentage is $\frac{216}{1200}\cdot 100=\boxed{18\%}.$