Correct Answer: $3\sqrt3:2$

Solution : Given: A circle is inscribed in an equilateral triangle and a square is inscribed in that circle.
Let the sides of the equilateral triangle be $a$ units, the radius of the circle be $r$ units and the sides of the square be $s$ units.
So, area of the equilateral triangle = $\frac{\sqrt3}{4}a^2$ sq. units
The radius of the circle inscribed in the triangle, $r$ = $\frac{a}{2\sqrt3}$ units
Length of the diagonal of the square inscribed in the circle = $\sqrt2s$
Since the diagonal of the square will be the diameter of the circle,
$\sqrt 2s=2r$
$⇒\sqrt 2s=2×\frac{a}{2\sqrt3}$
$⇒s=\frac{a}{\sqrt6}$ units
So, the area of the square = $s^2=\frac{a^2}{6}$ sq. units
$\therefore$ The required ratio = $\frac{\sqrt3}{4}a^2:\frac{a^2}{6}=3\sqrt3:2$
Hence, the correct answer is $3\sqrt3:2$.