Answer
$2.3\times 10^{30}Kg,1.2 Ms $
Work Step by Step
We can calculate the required mass as follows:
$T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt{Gm}}$
This can be simplified and rearranged as:
$m=\frac{4\pi^2 r^3}{T^2G}$
We plug in the known values to obtain:
$m=\frac{4\pi^2(10.5\times 10^9)^3}{(544320)^2(6.673\times 10^{-11})}=2.3\times 10^{30}Kg$
As we know that the mass of the sun is $1.99\times 10^{30}Kg$
Now the mass of HD68988 can be determined as
$\frac{2.31\times 10^{30}}{1.99\times 10^{30}}=1.2 \space times \space the\space mass\space of \space sun $
