Let the radius of larger circle be \(R\)
We know, a regular hexagon gets divided into 6 equilateral triangle when all of its vertices are joined with its centre.
Refer to the figure below;

In regular hexagon ABCDEEF, AB = BC = CD = DE = EF = FA = \(R\)

Now in isosceles triangle ABF, AB = AF = \(R\) and BF = \(\sqrt{3}R\)
Also notice that BF is the diameter of the smaller circle
Therefore, radius of the smaller circle = \(\frac{\sqrt{3}R}{2}\)

Ratio of the areas = \((\frac{\sqrt{3}}{2})^2 = \frac{3}{4}\)

Hence, option E