Answer:
time = distance / velocity = 2pr/v

Here r = 23,000 light years x 9.5 x 10¹⁵ m/ly = 2.2 x 10²⁰ m.

and the velocity is v= 220 km/s = 220,000 m/s
so we get for the time:

t = 2.0 x 3.1415 x 2.2 x 10²⁰ m / 220,000 m/s =  6.3 x 10 ¹⁵ seconds  or 6.3 x 10 ¹⁵ s / 3 x 10⁷ s/year = 2 x 10⁸ years.

b) How many times will the Sun orbit the galaxy in its 10 billion year lifetime?

This is easy! in 10 billion years how many times will the sun go around the galaxy, given that it finishes one circle in 0.2 x billion years?

N= 10 billion / 0.2 billion = 10/0.2 = 50 times.

The nearest "real" galaxy to us is the Andromeda Galaxy, also known as M 31. This galaxy has an apparent
size of 2 degrees (about four times the size of the full moon!). If its linear size is 25 kpc, how far away from us is it?

Looking from our perspective the Andromeda galaxy is 25 Kpc across, so half that distance is 12.5 Kpc.
The galaxy spans 2 degrees, so half that angle is 1 degree. (We divide the angle by 2 to get a right angle in this isoceles triangle).
Using trigonometry we get:
12.5 Kpc/ distance to the galaxy = tg (1 degree)
so
Distance to the galaxy = 12.5 Kpc / tg (1) = 720 Kpc = 2,200,000 Light years. (1 pc = 3.2 LY).

Problem #3:
 How long does it take for light from the galaxy in problem #2 to reach us?

Using the above result:
If the distance is 2,200,000 light years, that means that light takes 2,200,000 years to arrive from M31, the nearest galaxy!
That's more that two million years!

Problem #4: The visual magnitude of the bulge of M31 is 4.3. The spectral type of andromeda is G5, but for the purpose of this excersize lets assume that it is a G2 like our sun. Using the results of problem #3 find out the number of stars in the bulge of M31. For simplicity assume that ALL STARS are like our sun. Hint: Use box 24.1 in your book FETU to find the absolute magnitude of andromeda (its brightness if it was placed 10 parsecs from us). Compare that with the sun's magnitude, and then use the table to the left of figure 24.2 which tells the equivalence of magnitude difference and brightness
difference.

Answer:

Using the equation in box 24.1:
m-M = 5 log (r/10pc)
with the apparent magnitude m=4.3, and the distance to M31, from #2 above, r=720 Kpc = 720,000 pc

we get the absolute magnitude of:
M= - 20
which is very bright. Actually, that's 24.9 magnitudes brighter than the sun would be if placed at 10 pc (m=+4.79, from page 386 in your book).
When a star is 5 magnitudes smaller, it is 100 times brighter. If it is 10 magnitudes smaller, it is 100x100 times brighter. So 25 magnitudes smaller is 100⁵   times brigher, or 10¹⁰ times brighter.
So if we placed M31 bulge at 10pc. it would be 10 billion times brighter than the sun when placed at 10pc. That means that M31 is made of 10 billion stars like our sun! (if it is of the same spectral type.)

That is how we usually tell how many stars there are in a galaxy! (Can't count them. Astronomers don't get enough money to do that.)