Inscribed Triangle

Let $r$ be the radius of the circle. As shown in the image, the side length of the triangle is $r\sqrt{3}$ .

Plugging this into the formula for the area of an equilateral triangle, we get

$\dfrac{(r\sqrt{3})^{2}\sqrt{3}}{4}=48\sqrt{3}$

$\dfrac{3r^{2}}{4}=48$

$r^{2}=64$

Therefore, the area of the circle is $\boxed{64\pi}$ .