To determine the volume of the original cone, we first acknowledge that a plane parallel to the base cuts a cone into two pieces, both of which are similar to each other and to the original cone.
The original cone has a height of \( H = 27 \) cm. Let \( R \) represent the radius of the original cone's base. The smaller cone, removed by the cut, has a height of \( 18 \) cm and a base radius denoted by \( r \). Due to the similarity of the cones, their heights and radii are proportional: \(\frac{r}{R} = \frac{18}{27} = \frac{2}{3}\). Consequently, \( r = \frac{2}{3}R \).
We proceed to calculate the volumes. The volume of the original cone, \( V \), is given by:
\[ V = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi R^2 \times 27 = 9 \pi R^2 \]
The volume of the smaller cone, \( v \), is calculated as:
\[ v = \frac{1}{3} \pi r^2 \times 18 = \frac{1}{3} \pi \left(\frac{2}{3}R\right)^2 \times 18 = \frac{1}{3} \pi \times \frac{4}{9} R^2 \times 18 = \frac{24}{27} \pi R^2 = \frac{8}{9} \pi R^2 \]
The volume of the frustum resulting from the cut is the difference between the original cone's volume and the smaller cone's volume:
\[ V - v = 9 \pi R^2 - \frac{8}{9} \pi R^2 = \left(9 - \frac{8}{9}\right) \pi R^2 = \frac{81}{9} \pi R^2 - \frac{8}{9} \pi R^2 = \frac{73}{9} \pi R^2 \]
The problem statement provides that \( V - v = 225 \) cc:
\[ \frac{73}{9} \pi R^2 = 225 \]
Solving for \( \pi R^2 \):
\[ \pi R^2 = \frac{225 \times 9}{73} = \frac{2025}{73} \]
Therefore, the volume of the original cone is:
\[ V = 9 \pi R^2 = 9 \times \frac{2025}{73} = \frac{2025 \times 9}{73} = \frac{18225}{73} \approx 243 \text{ cc}\]
The volume of the original cone is \( \boxed{243} \) cc.