Stars more around COM. Distance of each star from COM is $ \large\frac{2}{3} \normalsize \times L \cos 30^{\circ}=\large\frac{L}{\sqrt 3}$

Force on each star M due to other two $=\large\frac{2GMM}{L^2}$$ \cos 30^{\circ}=\sqrt 3 \large\frac{GM^2}{L^2}$

acting along the radius This acts as the centripetal force

$\therefore \sqrt 3 \large\frac{GM^2}{L^2}=\frac{Mv^2}{\bigg(\Large\frac{L}{\sqrt 3}\bigg)^2}$

$v=\sqrt {\large\frac{GM}{L}}$

Total energy required to dismantle the system =Total KE +Total Gravitational energy

$\qquad= 3 \times \large\frac{1}{2} $$Mv^2+ \bigg[-3 \times \large\frac{GM^2}{L}\bigg]=\large\frac{3}{2} \frac{GM^2}{L}$

Hence a is the correct answer.