Answer :
To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.
In general terms the Orbital speed is described as,
[tex]V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}[/tex]
PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,
[tex]\frac{4\pi G\rho r}{3} = \frac{v^2}{r}[/tex]
[tex]\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}[/tex]
[tex]\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}[/tex]
[tex]\rho = \frac{1}{r^2}[/tex]
PART B) This time we have[tex]v=\omega t[/tex], where [tex]\omega[/tex] is the angular velocity (constant at this case)
[tex]\frac{4\pi G\rho r}{3} = \frac{v^2}{r}[/tex]
[tex]\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}[/tex]
[tex]\rho = \frac{3\omega r}{4\pi Gr}[/tex]
[tex]\rho = \frac{3\omega^2}{4\pi G} \propto constant[/tex]
PART C) If the total mass interior to any radius r is a constant,
[tex]\frac{4\pi G\rho r}{3} = \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r^2}=\frac{v^2}{r}[/tex]
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
[tex]v= \sqrt{\frac{1}{r}}[/tex]