Area of intersection of two circles

Let $AO = 2R$, $AC = \sqrt{2}R$, and $\angle AOB = 60^\circ$. In $\triangle AOD$, $AD = 2R \sin 30^\circ = R$. In right $\triangle ACD$, $\sin(\angle ACD) = \frac{AD}{AC} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}} \implies \angle ACD = 45^\circ$. So $\angle ACB = 90^\circ$. Area of common region = Area of major sector of smaller circle + Area of $\triangle ACB$ + Area of minor segment of larger circle $= \frac{270^\circ}{360^\circ} \pi (\sqrt{2}R)^2 + \frac{1}{2}(2R)(R) + \left[ \frac{60^\circ}{360^\circ} \pi (2R)^2 - \frac{\sqrt{3}}{4}(2R)^2 \right]$ $= \frac{3}{2}\pi R^2 + R^2 + \frac{2}{3}\pi R^2 - \sqrt{3}R^2 = \left(\frac{13\pi}{6} + 1 - \sqrt{3}\right)R^2$. Hence, option C is correct.