We can assume that the stars are equal in mass, and their orbits are circular
The orbital speed is 80000 m/s and at an orbital period of 10 months (or $2.628times 10^7$s) the length of the orbit is $2.1024times 10^{12}$ m or 14.05 AU The radius of the orbit therefore is $14.05/tau$ = 2.237AU.
The version of Keplers law given is $$T^2 =frac{a^3}{m_1+m_2}$$
substituting $T^2 =(10/12)^2 = 0.6944$ (divide by 12 to convert to years) and $a^3= 11.19$ gives $$m_1+m_2 = frac{11.19}{0.6944} = 16.12 mathrm{solar mass}.$$
Since $m_1=m_2$, the mass of each star is 8.06 solar masses, or $1.6times 10^{31}$kg
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