Area of the equilateral triangle is $\(\frac{\sqrt{3{4} \times\)$ Side $\(^2=7 \sqrt{3} \Rightarrow\)$ Side $\(=\sqrt{28}=2 \sqrt{7}\)$

The radius of the circle which circumscribes an equilateral triangle of side ' $\(a\)$ ' units is $\(\frac{a}{\sqrt{3\)$ (= circum-radius), so, the radius of the circle which would circumscribe an equilateral triangle of side $\(2 \sqrt{3}\)$ would be $\(\frac{2 \sqrt{7}}{\sqrt{3}}=2 \sqrt{\frac{7}{3}}\)$

So, the area of the circular table would be $\(\pi r^2=\frac{22}{7} \times\left(2 \sqrt{\frac{7}{3\right)^2=\frac{88}{3}\)$
Hence the answer is (B).