The particle to move in a circular path without chaging distance 'l' will happen if the resultant gravitational force on each particle provides the centripetal force necessary for the motion
$\large\frac{mv^2}{r}$$=\sqrt {F^2+F^2+2 F^2 \cos 60}$
$\large\frac{mv^2}{r}$$=\sqrt {3 }\; F$
Radius of the path is
$r= \large\frac{\sqrt 3}{2}$$l \times \large\frac{2}{3}$
$\quad= \large\frac{l}{3}$
$\therefore v= \sqrt {\large\frac {GM}{l}}$
Hence b is the correct answer.