Since, the centroid $G$ divide the line $AD$ in the ratio $2:1$

$\therefore AG=\frac{8}{3} and DG=\frac{4}{3}$
Here, $∆ABG$ is a right triangle, since $\angle AGB$ is $\frac{\pi }{2}$

In $∆ABG, \mathrm{tan }\frac{\pi }{3}=\frac{AG}{BG}$

$\Rightarrow BG=AG\mathrm{ cot }\frac{\pi }{3}$

$\Rightarrow BG=\frac{8}{3}\times \frac{1}{\sqrt{3}}=\frac{8}{3\sqrt{3}}$

$\text{Area of} ∆ADB=\frac{1}{2}\times AD\times BG$

$=\frac{1}{2}\times 4\times \frac{8}{3\sqrt{3}}=\frac{16}{3\sqrt{3}}$

Since, median divides a triangle into two triangles of equal area.

Therefore, area of $∆ABC=2\times \text{area of} ∆ADB$

$=2\times \frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}\mathrm{ }\mathrm{s}\mathrm{q}\mathrm{ }\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$