jmc

The original cube has $6$ square surfaces that each have an area of $3^2=9$, for a toal surface area of $6\cdot 9=54$. Since no two corner cubes touch, we can examine the effect of removing each corner cube individually. Each corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed. However, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$. Thus, the surface area does not change with the removal of a corner cube, and it remains $54$, which is answer $\boxed{54\text{ sq.cm}}$.