Difficult Astrophysics question, Need help please?

For a source emitting isotropically (same in all directions), with total power (luminosity) L, at a distance r, the power intensity (called "brightness" in this problem), I, is just L divided by the area of a sphere of radius r (because when it has travelled that distance in all directions, it has spread out evenly over that size sphere):

I = L/ 4πr^2

So knowing L and I, you get r from:

r = √(L/ 4πI)

This will come out in meters for the values given, so convert to lt-yr using:

1 lt-yr = c*yr

= 2.9979*10^8 m/s * 365.24*86400 s

= 9.46*10^15 m

L₀ = sun's luminosity

L₀ = 3.846e26 watts

m₀ = sun's apparent magnitude from Earth

m₀ = −26.74

F₀ = sun's luminous flux at Earth

F₀ = 1368 Wm⁻²

d₀ = 1 parsec

d₀ = 3.08568e16 meters

M = 4.83 − 2.5 log(L/L₀)

M = 71.293 − 2.5 log L

m − M = 5 log(d/d₀) − 5

m − M = 5 log d − 87.447

d = 10^(0.2m − 0.2M + 17.489)

d = 10^(0.2m + 0.5 log L + 3.23)

m = m₀ − 2.5 log(F/F₀)

d = 10^[0.2m₀ − 0.5 log(F/F₀) + 0.5 log L + 3.23]

d = 10^[ 0.5 log L − 0.5 log F − 0.55 ]

The equation just above uses MKS units for distance, luminosity, and luminous flux. Note that when L=L₀ and F=F₀ , then d is approximately equal to one astronomical unit. The error is one tenth of one percent.

Cepheid 1.

F = 9.3e-19 Wm⁻²

L = 3.9e30 W

d = 5.77e23 m = 18.7 Mpc

Cepheid 2.

F = 3.8e-19 Wm⁻²

L = 1.2e30 W

d = 5.01e23 m = 16.2 Mpc

Cepheid 3.

F = 8.7e-19 Wm⁻²

L = 2.5e30 W

d = 4.78e23 m = 15.5 Mpc

The "book value" for the distance to the M100 galaxy is 16.8 megaparsecs. So these numbers are definitely in the right ballpark.