Answer
The period is $~~0.71~years$
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:
$T = \sqrt{\frac{16\pi^2~R^3}{GM}}$
$T = \sqrt{\frac{16\pi^2}{(6.67\times 10^{-11}~N~m^2/kg^2)(1.99\times 10^{30}~kg)}~(7.50\times 10^{10}~m)^3}$
$T = 2.24\times 10^7~s$
$T = (2.24\times 10^7~s)(\frac{1~year}{365\times 24\times 3600~s})$
$T = 0.71~years$
The period is $~~0.71~years$
