One vertex of an equilateral triangle is (2,2) and its centroid is $\left(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$then length of its side is $k\sqrt{2}$ then k is 

The distance from the vertex and the centroid is $AG=\sqrt{{\left(2+\frac{2}{\sqrt{3}}\right)}^2+{\left(2-\frac{2}{\sqrt{3}}\right)}^2}=\sqrt{2\left(4+\frac{4}{3}\right)}=4\sqrt{\frac{2}{3}}$

We know that the height of the equailateral triangle is $\frac{6\sqrt{2}}{\sqrt{3}}$

Suppose that the length of the side is $a$

$$\begin{gathered} a^2-\frac{a^2}{4}=\frac{36·2}{3} \\ \frac{3a^2}{4}=24 \\ {a}^2=32 \\ a=4\sqrt{2} \end{gathered}$$