Answer : Option A

Explanation :

$MF#%\boxed{\text{Area of an equilateral triangle = }\dfrac{\sqrt{3}}{4}a^2 \\
\text{where a is length of one side of the equilateral triangle}}$MF#%

$MF#%\begin{align}
&\text{Area of the equilateral ” ABC = }\dfrac{\sqrt{3}}{4}a^2 =\dfrac{\sqrt{3}}{4}24^2 = 144\sqrt{3}\text{ cm}^2\text{.............(1)}
\end{align} $MF#%

$MF#%\boxed{\text{Area of a triangle = }\dfrac{1}{2}\text{bh}\\
\text{where b is the base and h is the height of the triangle}}$MF#%

$MF#%\begin{align}
&144\sqrt{3} = 36r\\\\
&\Rightarrow r = \dfrac{144}{36}\sqrt{3}= 4\sqrt{3}------------(3)
\end{align} $MF#%

$MF#%\boxed{\text{Area of a circle = }\pi r^2 \\ \text{ where = radius of the circle}}$MF#%

$MF#%\begin{align}
&\text{From (3), the area of the inscribed circle = }\pi r^2 = \pi \left(4\sqrt{3}\right)^2 = 48 \pi ------------(4)\\\\\\\\
&\text{Hence , Area of the remaining portion of the triangle }\\\\
&\text{Area of ” ABC “ Area of inscribed circle}\\\\
&144\sqrt{3} - 48\pi \text{ cm}^2
\end{align} $MF#%