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**Hint:**We are given the height h and both the radius r and R of a frustum and we need to find the volume of the frustum. The volume of the frustum is given by the formula $\dfrac{1}{3}h\left[ {\pi {R^2} + \pi {r^2} + \pi rR} \right]$.

Substituting the given values we get the required volume.

**Complete step by step solution:**

We are given that the height of the frustum is 6 cm

That is h = 6 cm

We know that a frustum has two radius r and R

Since we are given they are 5 cm and 9 cm

We have r = 5 cm and R = 9 cm

We are asked to find the volume of the frustum

The volume of the frustum is given by the formula

$ \Rightarrow \dfrac{1}{3}h\left[ {\pi {R^2} + \pi {r^2} + \pi rR} \right]$

Lets substitute the given values in this formula

We can have the pi as it is because the given options have a pi in them

$

\Rightarrow Volume = \dfrac{1}{3}\pi \left( 6 \right)\left[ {{{\left( 9 \right)}^2} + {{\left( 5 \right)}^2} + \left( 5 \right)\left( 9 \right)} \right] \\

\Rightarrow Volume = 2\pi \left[ {81 + 25 + 45} \right] \\

\Rightarrow Volume = 2\pi \left[ {151} \right] = 302\pi c{m^3} \\

$

**Hence we get the volume of the frustum to be $302\pi c{m^3}$**

Therefore the correct answer is option C.

Therefore the correct answer is option C.

**Note :**

A conical frustum is a frustum created by slicing the top off a cone (with the cut made parallel to the base). For a right circular cone, let be the slant height and and the base and top radii.

A right circular cone has only one frustum.

Frustum of a right circular cone is that portion of right circular cone included between the base and a section parallel to the base not passing through the vertex.

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