Here,

plz refer to the diagram attached,

First of all let us joined the AO , CO , BO, DO as they are all radii of the bigger circle = 3 (since radius = dia/2 = 6/2 = 3)

Let us join AC and BD, which are = 3 (since the diameter of the smaller circle = 3)

we find that △ AOC and △BOD are equilateral △ as all sides are equal and so all angles in the △'s are 60°

Now let us consider the sector AOC, we have the central angle = 60°

We know,

$\frac{Central angle}{360} = \frac{Sector Area AOC}{Circle Area}$

or $\frac{60}{360}= \frac{(Sector Area)}{(9\pi)}$

or Sector Area AOC = $\frac{(3\pi)}{2}$

Similarly Sector Area BOD = $\frac{(3\pi)}{2}$

Now we need the area of the △ AOB and △COD, in which we need the base of both the △'s i.e. AB and CD

Now let us consider the right angled △ BCD

we have BD =3 , CB = 6 , angle BDC =90° , angle DBC = 60° and angle BCD = 30° (angle BDC = 90° because the line AB is parallel to line CD and both are inscribed within a circle)

SO the triangle is a 30-60-90 triangle and the sides are in the ratio of $1: \sqrt{3} : 2$

or we can write as $3 : 3\sqrt{3} : 6$ ; such that the side $CD = 3\sqrt{3}$

Now the Area of the △COD,

$\frac{1}{2} * base * altitude = \frac{1}{2} * CD * \frac{3}{2}$ (Altitude = 1.5 i.e. the radius of the smaller circle)

= ${\frac{1}{2} * 3\sqrt{3} * \frac{3}{2}}$

i.e. Area of the △COD = $\frac{9\sqrt3}{4}$

Similarly in △ AOB the area = $\frac{9\sqrt3}{4}$

Now the area of the shaded region = Area of the sector AOC + Area of the sector BOD + Area of the △ AOB + Area of the △ COD = $\frac{(3\pi)}{2} + \frac{(3\pi)}{2} + \frac{9\sqrt3}{4} + \frac{9\sqrt3}{4}$ = $3\pi + \frac{9\sqrt3}{2}$