Answer
See the detailed answer below.
Work Step by Step
a) As we see in the figure below, $m_1=2\times 10^{30}$ kg and $m_2=6\times 10^{30}$ kg. We need to find the center of mass of the system, so we can find its periodic time for one full rotation.
$$x_{cm}=\dfrac{m_1R_1 +m_2R_2 }{m_1+m_2}$$
We chose $m_1$ to be our origin, so $R_1=0$ and $R_2=d$
$$x_{cm}=\dfrac{m_1(0) +m_2d }{m_1+m_2}$$
Plugging the known;
$$x_{cm}=\dfrac{ (6\times 10^{30}) (2\times 10^{12}) }{(2\times 10^{30})+(6\times 10^{30})}=\bf 1.5\times 10^{12}\;\rm m$$
Now we know that the system is rotating around its center of mass as shown below.
The radius of the circular path of star 1 is $R_1=1.5\times 10^{12}\;\rm m$ while the radius of the circular path of star 2 is $R_2=d-R_1=0.5\times 10^{12}\;\rm m$. See the figures below
The two stars complete one full circle in the same time; imaging it as a rod rotating at some point closer to its right end.
The neft force exerted on star 1 is given by
$$\sum F_1=F_{G}=m_1a_{r}$$
where $a_r$ is the centripetal acceleration and is given by $v^2/r$ and $F_G$ is the gravitational force exerted by star 2 on 1.
$$\dfrac{G \color{red}{\bf\not}m_1m_2}{d^2}=\dfrac{ \color{red}{\bf\not}m_1v_1^2}{R_1}$$
Recalling that $v=2\pi r/T$;
$$\dfrac{G m_2}{d^2}=\dfrac{ \left(\frac{2\pi R_1}{T}\right)^2}{R_1}$$
$$\dfrac{G m_2}{d^2}=\dfrac{ 4\pi^2R_1^{ \color{red}{\bf\not}2} }{T^2 \color{red}{\bf\not}R_1}$$
$$\dfrac{Gm_2}{d^2}=\dfrac{ 4\pi^2R_1 }{T^2 }$$
Hence, the periodic time is given by
$$T =\sqrt{\dfrac{ 4\pi^2R_1d^2 }{ Gm_2}}$$
Plugging the known;
$$T =\sqrt{\dfrac{ 4\pi^2(1.5\times 10^{12})(2\times 10^{12})^2 }{ (6.67\times 10^{-11})(6\times 10^{10})}}$$
$$T=\bf 7.693\times 10^{18}\;\rm s\approx \color{red}{\bf2.44\times 10^{11}}\;\rm yr$$
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b) The speed of star 1 is
$$v_1=\dfrac{2\pi R_1}{T}=\dfrac{2\pi (1.5\times 10^{12})}{7.693\times 10^{18}}$$
$$v_1=\color{red}{\bf 1.22511\times10^{-6}}\;\rm m/s$$
b) The speed of star 2 is
$$v_2=\dfrac{2\pi R_2}{T}=\dfrac{2\pi (0.5\times 10^{12})}{7.693\times 10^{18}}$$
$$v_2=\color{red}{\bf 4.0837\times10^{-7}}\;\rm m/s$$
