Answer
$\omega = 2.2 \times 10^{-7}~rad/s$
Work Step by Step
Let $R$ be the orbital radius of each star. The gravitational force provides the centripetal force for each star. We can find an expression for the period:
$\frac{GM^2}{(2R)^2} = \frac{Mv^2}{R}$
$\frac{GM^2}{(2R)^2} = \frac{M(2\pi~R/T)^2}{R}$
$\frac{GM^2}{4R^2} = \frac{4\pi^2~M~R}{T^2}$
$T^2 = \frac{16\pi^2~R^3}{GM}$
$T = \sqrt{\frac{16\pi^2~R^3}{GM}}$
We can find the period of each star:
$T = \sqrt{\frac{16\pi^2~R^3}{GM}}$
$T = \sqrt{\frac{(16\pi^2)(1.0\times 10^{11}~m)^3}{(6.67\times 10^{-11}~N~m^2/kg^2)(3.0\times 10^{30}~kg)}}$
$T = 2.8\times 10^7~s$
We can find the angular speed:
$\omega = \frac{2\pi~rad}{2.8\times 10^7~s}$
$\omega = 2.2 \times 10^{-7}~rad/s$
