If the smaller cone's volume is $\frac{8}{125}$ of the original cone's volume, take a cube root to find the factor relating corresponding lengths.

The frustum has a lateral surface, a larger circular base, and a smaller circular top face. The lateral surface is the original cone's lateral area minus the removed cone's lateral area.

An algebraic similarity approach is faster here. For a quick verification, enter r=15, h=20, and l=sqrt(r^2+h^2) to confirm the original slant height.

Enter k=(8/125)^(1/3). Desmos gives $k=0.4$, which is $\frac{2}{5}$. The area scale factor is k^2.

Enter (1-k^2)*pi*r*l+pi*r^2+pi*(k*r)^2. This adds the frustum's lateral area, lower base area, and upper base area.

Since the base area is $225\pi$, the radius satisfies $\pi r^2=225\pi$, so $r=15$.

The vertical height is $20$, so the slant height is

For similar cones, volume scales by the cube of the linear scale factor. Therefore,

The linear scale factor is $\frac{2}{5}$. Thus, all corresponding areas of the smaller cone are $\left(\frac{2}{5}\right)^2=\frac{4}{25}$ of the corresponding areas of the original cone.

The original cone's lateral area is

The removed small cone has lateral area $\frac{4}{25}(375\pi)=60\pi$. Therefore, the frustum's lateral area is

The lower base of the frustum has area $225\pi$. Its upper base is the cut face, whose area is $\frac{4}{25}(225\pi)=36\pi$.

Thus, the total surface area is