This problem involves a cylinder inscribed within a cone. To find the volume of the cylinder, we need its radius and height. We are given the radius of the cylinder, but we need to find its height using similar triangles.
1. Understand the Dimensions of the Cone:
- The image shows the total height of the cone $(H)$ is 8 units.
- The diameter of the base of the cone is 12 units, so its radius $\((R)$ is $12 / 2=6\)$ units.
2. Understand the Dimensions of the Cylinder:
- The radius of the inscribed cylinder $(r)$ is given as 3 units.
- Let the height of the inscribed cylinder be $h$.
3. Use Similar Triangles:

Imagine a cross-section of the cone and cylinder. This forms two similar right triangles:
- The large triangle formed by the cone's height $(H)$, its radius $(R)$, and its slant height.
- The smaller triangle formed by the cone's apex, the cylinder's radius $(r)$, and the part of the cone's height above the cylinder.

Let the height of the cone above the cylinder be $H^{\prime}$.
The base of this smaller triangle is the cylinder's radius $(r)$.
The ratio of corresponding sides in similar triangles is equal:

$$
\(\frac{H}{R}=\frac{H^{\prime}}{r}\)
$$


Substitute the known values:

$$
\(\frac{8}{6}=\frac{H^{\prime}}{3}\)
$$


Now, solve for $H^{\prime}$ :

$$
\(\begin{aligned}
& 8 \times 3=6 \times H^{\prime} \\
& 24=6 H^{\prime} \\
& H^{\prime}=\frac{24}{6}=4 \text { units. }
\end{aligned}\)
$$

4. Find the Height of the Cylinder ( $h$ ):

The height of the cylinder ( $h$ ) is the total height of the cone ( $H$ ) minus the height of the cone above the cylinder ( $\(H^{\prime}\)$ ).

$$
\(\begin{aligned}
& h=H-H^{\prime} \\
& h=8-4=4 \text { units. }
\end{aligned}\)
$$

5. Calculate the Volume of the Cylinder:

The formula for the volume of a right circular cylinder is $\(V=\pi r^2 h\)$.
Substitute the cylinder's radius $\((r=3)\)$ and height $\((h=4)\)$ :

$$
\(\begin{aligned}
& V=\pi(3)^2(4) \\
& V=\pi(9)(4) \\
& V=36 \pi \text { cubic units. }
\end{aligned}\)
$$


The final answer is $\(36 \pi\)$.