$\text{2 }\angle \text{B}={\text{sin}}^{-1}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)+{\text{sin}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

      $={\text{tan}}^{-1}\left(\frac{\sqrt{6}+1}{\sqrt{3}-\sqrt{2}}\right)+{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

     $={\text{tan}}^{-1}\left(\frac{\frac{\sqrt{6}+1}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{2}}}{1-\frac{\sqrt{6}+1}{\sqrt{3}-\sqrt{2}}·\frac{1}{\sqrt{2}}}\right)$

     $={\text{tan}}^{-1}\left(\frac{\left(\sqrt{6}+1\right)\sqrt{2}+\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-\left(\sqrt{6}+1\right)}\right)$

     $={\text{tan}}^{-1}\left(\frac{\sqrt{12}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{\sqrt{6}-2-\sqrt{6}-1}\right)$

     $={\text{tan}}^{-1}\left(\frac{2\sqrt{3}+\sqrt{3}}{-3}\right)$

     $={\text{tan}}^{-1}\left(\frac{3\sqrt{3}}{-3}\right)$

    $={\text{tan}}^{-1}\left(-\sqrt{3}\right)$

      $=\frac{2\pi }{3}$

According to question

$\angle \text{A}=\angle \text{B}=\frac{1}{2}\left({\text{sin}}^{-1}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)+{\text{sin}}^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$

$\Rightarrow \angle \text{A}=\angle \text{B}=\frac{1}{2}\left(\frac{2\pi }{3}\right)$

$\Rightarrow \angle \text{A}=\angle \text{B}=\frac{\pi }{3}$

$\therefore \angle \text{C}=\pi -\angle \text{A}-\angle \text{B}$

        $=\pi -\frac{2\pi }{3}$

       $=\frac{\pi }{3}$

Since $\Delta \text{ABC}$ is an equilateral $\Delta$

$\therefore \text{ ar }\left(\Delta \text{ABC}\right)=\frac{\sqrt{3}}{4}{\text{a}}^2$

                       $=\frac{\sqrt{3}}{4}{\text{c}}^2$

                        $=\frac{\sqrt{3}}{4}·{\left(6\cdot 3^{1/4}\right)}^2$

                      $=\frac{\sqrt{3}}{4}\sqrt{3}\cdot 6\times 6$

                      $=3\times 3\times 3$

                      $=27$