Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube.  The surface area of the remaining figure is
[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0));  draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99));  draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]
$\text{(A)}\ 19\text{ sq.cm} \qquad \text{(B)}\ 24\text{ sq.cm} \qquad \text{(C)}\ 30\text{ sq.cm} \qquad \text{(D)}\ 54\text{ sq.cm} \qquad \text{(E)}\ 72\text{ sq.cm}$
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 $\boxed{D}$.