We know F = L/(4 * pi * r^2) for any luminous source.  For a source
emitting thermal radiation, we also know F = (sigma) * t^4.  The Sun, in
its current main sequence phase as well as its red giant and white dwarf
phases, is a luminous source emitting thermal radiation; so for the Sun,
L = (4 * pi * r^2) * (sigma) * T^4  in any phase. 


(a) We can set up a ratio where

L(main seq)    4 * pi * (sigma) * r^2 * T^4   (7.0 * 10^8 m)^2 * (5800 K)^4
-----------  = ---------------------------- = ------------------------------
L(red giant)   4 * pi * (sigma) * r^2 * T^4   (1.2 * 10^11 m)^2 * (3000 K)^4

  = (0.00583)^2 * (1.93)^4 = 0.000476

The red giant Sun = 1/0.000476 = 2100 times more luminous than the Sun is now.


(b) We can set up a similar ratio where

L(main seq)    4 * pi * (sigma) * r^2 * T^4   (7.0 * 10^8 m)^2 * (5800 K)^4
-----------  = ---------------------------- = ------------------------------
L(wh dwarf)   4 * pi * (sigma) * r^2 * T^4    (6.4 * 10^6 m)^2 * (40000 K)^4

  = (109)^2 * (0.145)^4 = 5.25

The white dwarf is 1/5.25 = 0.19 times as luminous as the Sun is now.


(c) F = L/(4 * pi * r^2), where in this case r = the distance from
the surface of the Earth to the surface of the Sun.  In the case of
the main sequence and white dwarf, the Sun's radius is much smaller
than that distance, so we can approximate r = 1.5 * 10^11 m, or 1 AU.

F(main seq) = 3.9 * 10^26 W / (4 * pi * (1.5 * 10^11 m)^2 )
            = 1400 W/m^2

F(wh dwarf) = 0.19 * 3.9 * 10^26 W / (4 * pi * (1.5 * 10^11 m)^2 )
            = 262 W/m^2

But for the red giant, the radius is a significant fraction of 1 AU,
so we must subtract this radius from 1 AU to get "r" in the equation:
r = (1.5 * 10^11 m) - (1.2 * 10^11 m) = 0.3 * 10^11 m.  So,

F(red giant) = 2100 * 3.9 * 10^26 W / (4 * pi * (0.3 * 10^11 m)^2 )
             = 7.2 * 10^7 W/m^2

So during the red giant phase, we get over 50,000 times the flux
from the Sun that we'd normally get.  Some serious global warming!