The Stefan-Boltzmann Law was first discovered experimentally in 1879 by Stefan, then derived theoretically in 1884 by his student Boltzmann. It says:
Note the strong dependence on temperature: -- as the fourth power of the temperature. I.e., double the temperature, and the radiated power will increase by a factor 24 = 16. Triple the temperature, and the power will increase by a factor 34 = 81, etc. The Stefan-Boltzmann Law is valid only for perfect radiators (called "blackbodies"). Actual radiating surfaces are not perfect radiators, and will always radiate less than the luminosity given by the S-B Law -- typically some 10 - 80%. But the radiating surfaces (called "photospheres") of stars are fairly good approximations to black bodies, and typically radiate a luminosity of more than 90% of the value given by the S-B Law.
The value of the Stefan-Boltzmann constant, s = 5.67 x 10-5 ergs cm-2 s-1 K-4, is determined by experiments in Earth labs. We assume that the S-B Law, with the same value of s, is valid everywhere in the universe. That assumption is one example of the Principle of Universality of Physical Laws.
1. The Earth: The average temperature of the Earth's surface must be about 10 C. We must first convert this to the Kelvin scale: T(K) = T(C) + 273 = 283 K. The Earth's surface area is about A = 4p R2 = 4p (6.4 x 108 cm)2 = 5.15 x 1018 cm2. Now we can calculate the radiated luminosity of the Earth: L = 5.15 x 1018 x 5.67 x 10-5 x (283)4 = 1.87 x 1024 ergs s-1. To change ergs s-1 to Watts, multiply by 10-7. So, L = 1.87 x 1017 Watts. (Actually, the Earth radiates only about half this much luminosity into outer space because it is not a black body.)
2. The Sun: The temperature of the Sun's photosphere is 5800 K and its radius is 7 x 1010 cm. So the Stefan-Boltzmann Law gives a luminosity L = 4 x 1033 erg s-1 = 4 x 1026 Watts.
3. Your body: (This calculation will be a bit more subtle because you must correct for the temperature of the environment.)
Your body temperature is about T(F) = 98.6 F. To use the S-B Law, we must convert this to the Kelvin scale. To do this, first convert your body temperature to Centigrade: T(C) = (5/9)[T(F) -32] = 37 C. Then, to convert this to Kelvin, add 273: T(K) = T(C) + 273 = 310 K. Now, let's estimate your body surface area. Approximate your body shape roughly by a rectangular block, of height 180 cm, width 30 cm, and depth 15 cm, giving a surface area A = 2[180 x 30 + 180 x 15 + 30 x 15] = 17,100 cm2. Now we can calculate the luminosity of your body: L = 17,100 x 5.67 x 10-5 x (310)4 = 8.95 x 109 ergs s-1. So, according to the S-B Law, your body could radiate a luminosity L = 895 Watts. That's a lot! But this would only be true if your body was in outer space where the temperature of the sky is very cold -- about 5 K -- and if you were naked. If you wore an insulating space suit, you would radiate much less than this because the surface temperature of the suit would be much colder than your body. On Earth, you won't lose nearly as much heat as this, because your body is absorbing heat from its surroundings at the same time it is radiating. To correct for this absorption, you should use a modified Stefan-Boltzmann Law:
L = As (T4 - Tenv4)
Where Tenv is the temperature of the environment. Let's say that Tenv = 290 K (about 63 F). Using this formula, the calculation gives L = 210 Watts. But, if you put your clothes on, you can cut this radiation down by a big factor -- say, a factor 4 with light clothes and a sweater. Then you are losing about 50 Watts, the equivalent of about 1 light bulb.