Calculating Temperature Inside Greenhouse According to this forum I can calculate the internal temperature of a greenhouse with the Stefan-Boltzmann Law via the following variables:
Outside Temperature 273.15K
Solar Radiation 
1000 watts per square meter.
Area Ground
2.70 m^2
Stefan-Boltzmann Constant
σ = 5.670367(13)×10−8 W m−2 K−4

As you can see, I get 390 K as the greenhouse's internal temperature, which is obviously wrong. What's at fault here?
 A: Although the approach you are taking to predicting the equilibrium greenhouse temperature isn't a half-bad start on the matter, there are quite a number of things wrong (both theoretically and practically) with your starting assumptions. 
First off, the number you quote of 1000 Watts/square meter isn't altogether wrong, but we need to understand that this number is only applicable if the Sun is directly overhead.  Obviously, you won't be receiving that same amount of irradiance 10 minutes before sunset, and you won't be receiving any at all after dark.  My point in all of this is that the 1000 Watts/m^2 is oversimplified somewhat.
The second oversimplification I see in your calculations is that you are using $A_{ground}$ in some incorrect ways.  It is true that if the sun were straight overhead, the $A_{ground}$ would be used to calculate the total power coming in to the greenhouse from the sunlight.  However, you need to use $6A_{ground}$ when calculating the surface area of the greenhouse (assuming a cubic greenhouse) to determine the total power radiated out by the greenhouse in the form of blackbody radiation.  In your calculations, I don't see anywhere that you have done this.
At any rate, here's how I would go about solving this problem:
$A_s$ : the area of one face of the (cubic) greenhouse
$J_{sun} = 1000 W/(m^2)$ : the amount of solar power incident on the top surface of the greenhouse
$T_{atm}$ = 273.15 K is the atmospheric temperature
$T_g$ is the equilibrium temperature of the greenhouse (the answer to our question), once it has risen to a temperature where the power being radiated out is equal to the power being radiated in.
So at any rate, assuming:
1.) There is no energy transfer other than sunlight coming in, blackbody radiation from the atmosphere coming in, and blackbody radiation escaping from the glass of the greenhouse.
2.) The greenhouse absorbs all of the sunlight that hits it (i.e. the windows have no reflectivity.
3.) There is no conductive heat loss to the atmosphere, ground, or convective heat loss (hot air escaping out of the top)

we can start with: 

Power in = Power out
  Power in from sunlight + Power in from atmospheric blackbody radiation = Power radiated out from greenhouse
  Now:
  $J_{sun} * A_{g} + J_{atm} * 6*A_{g} = J_g * 6 * A_{g}$
  $1000 * 2.70 + \sigma * (T_{atm})^4 * 6 * 2.70 = \sigma * (T_g)^4 * 2.70$
  plugging in the $\sigma$ you gave, and plugging in the atmospheric temperature you gave, and solving for $T_g$ gives me:
  $T_g = 239 K$

Obviously this is way too cold.  Conduction of heat coming in from the atmosphere and conduction of heat going out from the greenhouse need to be taken into consideration as well in order to get this into a more realistic range.  In order to refurbish this whole calculation, we need to know how well the glass insulates heat, how well the atmosphere conducts air, temperature of ground, conductivity of ground, etc.  I'll leave all that for some other exercise.
Hope this helps.
A: I think the trouble with your calculation is that your beginning equation says: Solar radiation incident on the ground = Energy being emitted by the ground (by virtue of its temperature), which implies that ground is absorbing all the radiation incident on it. Left side of the equation should be multiplied by absorptivity of the ground to solar radiation.
A: It is way too hot because you don't take into account conduction loss through the glass.
For surface radiation, not sure (that is why I posted here) but outside glass should radiate $\epsilon \sigma T_{out}^4 (A_t - A_g)$, with $t$ and $g$ stand for total surface and ground surface, $\epsilon$ emissivity of glass.
Then inside the radiation is complicated because of the view factors from each surface but roughly speaking we have inside:
$\sigma T_{in}^4 ( \epsilon (A_t-A_g) + \epsilon_g A_g )$
With $\epsilon_g$ ground emissivity. So the power balance can be written:
$$ \sigma T_{in}^4 ( \epsilon (A_t-A_g) + \epsilon_g A_g ) - \epsilon \sigma T_{out}^4 (A_t - A_g) - \frac{\lambda (A_t-A_g)}{h} (T_{in} - T_{out}) + \Phi A_g  = 0 $$
with $\Phi$ sun's power in W/m$^2$, $h$ glass thickness.
