A

Step 1: According to Kepler's Third Law of planetary motion, in a binary star system, the ratio of the squares of the periods of the two stars is equal to the ratio of the cubes of the semi-major axes of their orbits. In this case, we can express the relationship as:

$$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$$
where $T_1$ and $T_2$ are the orbital periods of stars α and β, respectively, and $r_1$ and $r_2$ are their respective orbital radii.

Step 2: We make use of Newton's law of gravitation, which states:
$$F = G \frac{m_1 m_2}{r^2}$$
For circular orbits, the centripetal force is also given by:
$$F = \frac{m v^2}{r}$$

Thus, we have:
$$\frac{m_α v_α^2}{r_α} = G \frac{m_α m_β}{(r_α + r_β)^2}$$
and
$$\frac{m_β v_β^2}{r_β} = G \frac{m_α m_β}{(r_α + r_β)^2}$$

Step 3: From the relationship of the radii, we set $r_α = 1.00 \times 10^9$ km and $r_β = 5.00 \times 10^8$ km. The total distance $r_1 + r_2 = (1.00 + 0.5) \times 10^9 = 1.5 \times 10^9$ km.

Step 4: The gravitational force can be derived to find the ratio of the masses:
$$\frac{m_β}{m_α} = \frac{r_α^3}{r_β^3}$$ which gives:

Step 5: Now substituting in the values:

$$\frac{m_β}{m_α} = \frac{(1.00 \times 10^9)^3}{(5.00 \times 10^8)^3} = \frac{1.00^3}{0.5^3} = \frac{1}{0.125} = 8$$

Therefore, the ratio of the masses of star β to star α is:
Answer: 8.