Answer
$\text{see the solution below}$
Work Step by Step
in this problem we use the given density and the Sun’s mass to calculate the size of the Sun.
$$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi r_{\mathrm{Sun}}^{3}} \rightarrow$$
$$r_{\mathrm{Sun}}=\left(\frac{3 M}{4 \pi \rho}\right)^{1 / 3}=\left[\frac{3\left(1.99 \times 10^{30} \mathrm{kg}\right)}{4 \pi\left(10^{-26} \mathrm{kg} / \mathrm{m}^{3}\right)}\right]^{1 / 3}$$
$$= 3.62 \times 10^{18} \mathrm{m}\left(\frac{1 \mathrm{ly}}{9.46 \times 10^{15} \mathrm{m}}\right)=382 \mathrm{ly} \approx[400 \mathrm{ly}]$$
and we continue our solution as following:
$$\frac{r_{\mathrm{Sun}}}{d_{\mathrm{Earth}-\mathrm{Sun}}}=\frac{3.62 \times 10^{18} \mathrm{m}}{1.50 \times 10^{11} \mathrm{m}} \approx \frac{\sqrt{2 \times 10^{7}}}{d_{\mathrm{galaxy}}}= $$
$$\frac{382 \mathrm{ly}}{100,000 \mathrm{ly}} \approx \left[4 \times 10^{-3} \right]$$
