Answer
$\approx 10\;\rm yr$
Work Step by Step
We need to find the centripetal acceleration that makes the three stars remain in their places relative to each other, as we see in the figure below.
This could be found if we found the net force exerted on one of these 3 stars and then found its period of rotation.
Note that the 3 stars are rotating as one unit so the period of one star is the period of the two others.
Now we need to find the net radial forces exerted on star 3. The figure below shows that the net force component in the direction perpendicular to the radial direction is zero.
$$\sum F_{r3}=F_{G,\rm 1on3}\cos30^\circ+F_{G,\rm 2on3}\cos30^\circ=m a_r$$
The three stars have the same mass;
$$\dfrac{Gm \color{red}{\bf\not}m }{L^2}\cos30^\circ+\dfrac{Gm \color{red}{\bf\not} m }{L^2}\cos30^\circ= \color{red}{\bf\not}m\dfrac{v^2}{r}$$
where $r$ is the radius of the circular path of any star.
$$ \dfrac{2Gm }{L^2}\cos30^\circ= \dfrac{v^2}{r}$$
where $v=2\pi r/T$
$$ \dfrac{2Gm }{L^2}\cos30^\circ= \dfrac{4\pi^2r^{ \color{red}{\bf\not}2}}{ \color{red}{\bf\not}rT^2}$$
Hence,
$$T^2= \dfrac{4\pi^2L^2r}{2Gm\cos30^\circ }$$
where, from the geometry of the figure below, $\cos30^\circ=\dfrac{1}{2}L/r$;
Thus,
$$T^2= \dfrac{4\pi^2L^2\left(\dfrac{L}{2\cos30^\circ}\right)}{2Gm\cos30^\circ }$$
$$T =\sqrt{ \dfrac{4\pi^2L^3 }{4Gm(\cos30^\circ )^2 }}$$
Plugging the known;
$$T =\sqrt{ \dfrac{4\pi^2(1\times10^{12})^3 }{4(6.67\times10^{-11})(1.99\times 10^{30})(\cos30^\circ )^2 }}$$
$$T=3.15\times10^8\;\rm s\approx \color{red}{\bf 10}\;\rm yr$$
Therefore, the period of rotation around the system's center of mass is about 10 years.
