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Let us draw the figure:

Here, green circle indicates the circumcircle with radius R and orange circle indicates the incircle with the radius r.

We construct a perpendicular arm OP at the base BC of the triangle ABC. OB is the angle bisector of the $\angle ABC$ and as it is an equilateral triangle so every angle must be equal to 60°. Therefore, the angle OBP will be 30°.

Let us first calculate the radius of the circumcircle. The circumference of a circle is given by 2$\pi $r. let R be the radius of the circumcircle and its circumference is given as $24\pi $units.

Therefore, 2$\pi $R = $24\pi $

$ \Rightarrow $2R = 24

$ \Rightarrow $R= 12 units.

Now, for the radius r of the incircle: in triangle ABC,

Sin B = $\dfrac{{perpendicular}}{{hypotenuse}}$

$ \Rightarrow $$\sin {30^ \circ } = \dfrac{r}{R}$

$

\Rightarrow \dfrac{1}{2} = \dfrac{r}{R} \\

\Rightarrow r = \dfrac{R}{2} \\

\Rightarrow r = \dfrac{{12}}{2} = 6 \\

$

Therefore, the radius of the incircle is found to be 6 units.

Now, we can calculate the area of incircle using the formula of the area of the circle: $\pi {r^2}$

$

\Rightarrow area(incircle) = \pi {r^2} = \pi \times {(6)^2} \\

\Rightarrow area(incircle) = 36\pi \\

$