Figure I shows a piece of wood is in the form of a right circular cone with a base diameter of 18 cm.

1. Find the radius from the given diameter.

The base diameter is $18\text{ cm}$, so
$r=\frac{18}{2}=9\text{ cm}.$

2. Use the curved surface area of a cone to find the slant height.

Curved surface area of a cone is $\pi rl$.
$\pi(9)l=135\pi$
$9l=135$
$l=15\text{ cm}.$

3. Use Pythagoras' theorem to find the height of the cone.

In the right triangle formed by the radius, height and slant height,
$h^2+9^2=15^2$
$h^2=225-81=144$
$h=12.$

1. Find the height of the upper small cone using similar cones.

The upper cone and the original cone are similar.

Diameters are in the ratio
$6:18=1:3.$

So their heights are also in the ratio $1:3$.

Original cone height $=12\text{ cm}$ from part (a), hence
$\text{height of upper cone}=\frac{1}{3}\times 12=4\text{ cm}.$

2. Subtract the upper cone height from the original cone height to find the frustum height.

The frustum height is
$x=12-4=8.$

1. Find the volume of the frustum using the radii and the height from part (b).

For the frustum, the height is $x=8\text{ cm}$, the larger radius is $9\text{ cm}$ and the smaller radius is $3\text{ cm}$.

$\text{Volume of frustum}=\frac{1}{3}\pi h(R^2+Rr+r^2)$

$=\frac{1}{3}\pi(8)(9^2+9\times3+3^2)$

$=\frac{8}{3}\pi(81+27+9)$

$=312\pi\text{ cm}^3$

2. Subtract the volume of the cylindrical hole to find the remaining volume of wood.

The cylindrical hole has diameter $6\text{ cm}$, so its radius is $3\text{ cm}$ and its height is $8\text{ cm}$.

$\text{Volume of cylinder}=\pi r^2h$

$=\pi(3^2)(8)=72\pi\text{ cm}^3$

Remaining volume:

$312\pi-72\pi=240\pi\text{ cm}^3$

3. Convert the density to grams per cubic centimetre and calculate the mass remaining.

Since $1\text{ m}^3=1\,000\,000\text{ cm}^3$ and $810\text{ kg}=810\,000\text{ g}$,

$810\text{ kg/m}^3=\frac{810\,000}{1\,000\,000}=0.81\text{ g/cm}^3$

Mass remaining:

$240\pi\times0.81=194.4\pi$

$=610.7\ldots\text{ g}$

$\approx 611\text{ g}$