What will be the circumference of circle in this question?

Let the centers of the lower left vertex circle be $O_1$, horizontally next to it is the circle with center $O_2$, horizontally next to the latter is circle $O_3$, then we go analogously up to denote the centers of the next two consecutive circles by $O_4$ and $O_5$ and the center of the last circle is $O_6$, i.e. the center of the second, different from $O_2$, circle tangent to the circle with center $O_1$. The vertex circles have centers $O_1, O_3, O_5$ forming an equilateral triangle and the points $O_2, O_4, O_6$ are the midpoints of the corresponding edges of the triangle $O_1O_3, O_3O_5, O_5O_1$. Therefore the three segments $O_1O_4, \, O_3O_6, \, O_5O_2$ intersect at the centroid $O$ of the triangle $O_1, O_3, O_5$. The centroid $O$ is the center of the large circle tangent to the three circles with centers $O_1, O_3, O_5$. The radius $R$ of that large tangent circle is $R=|OO_1|+1$. As $O$ is the centroid of the equilateral triangle $O_1, O_3, O_5$ it follows that $|OO_1|=\frac{2}{3}|O_1O_4|$ while $|O_1O_4|=\frac{\sqrt{3}}{2}|O_1O_3|=\frac{\sqrt{3}}{2}4=2\sqrt{3}$. Thus $|OO_1|=\frac{2}{3}2\sqrt{3}= \frac{4}{\sqrt{3}}$ and therefore $R=\frac{4}{\sqrt{3}}+1=\frac{4+\sqrt{3}}{\sqrt{3}}$. The circumference of the large tangent circle is $2\large(\frac{4+\sqrt{3}}{\sqrt{3}}\large)\pi$.