Background
The specific intensity or brightness, $I_{\nu}$, is defined as:
$$
I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}
$$
where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency.
We can define a net flux at each frequency as:
$$
F_{\nu} = \int d\Omega \ I_{\nu} \ \cos{\theta} \tag{2}
$$
where we can get a total flux by integrating $F_{\nu}$ over all $\nu$.
For blackbody radiation, $I_{\nu}$ goes to the Planck function, $B_{\nu}$, defined as:
$$
B_{\nu} \left( T \right) = \frac{ 2 \ h \ \nu^{3} }{ c^{2} } \left[ e^{h \ \nu/k_{B} T} - 1 \right]^{-1} \tag{3}
$$
where $k_{B}$ is the Boltzmann constant, $c$ is the speed of light, $h$ is the Planck constant, and $T$ is the temperature.
According to the Stefan–Boltzmann law, the energy emitted per unit time per unit area, $\mathcal{F}$, is defined as:
$$
\begin{align}
\mathcal{F} & = \pi \ \int \ d\nu \ B_{\nu} \left( T \right) \tag{4a} \\
& = \sigma \ T^{4} \tag{4b}
\end{align}
$$
where $\sigma$ is the Stefan-Boltzmann constant.
If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?
You are close in your assumption. The ratio of the emergent flux (i.e., Equations 4a,b) before to after would be $\propto N^{-4}$. However, the intensity at any given frequency and temperature is given by $B_{\nu} \left( T \right)$ (i.e., Equation 3).
For instance, if we define $T_{1}$ = 1 K, $T_{2}$ = 1/5 K, and $\nu_{o} = 10^{11}$ Hz, then the ratio $B_{\nu_{o}} \left( T_{1} \right)/B_{\nu_{o}} \left( T_{2} \right)$ would be $\sim 2.2 \times 10^{8}$, which is much larger than $5^{4}$ (= 625), where we have:
$$
B_{\nu_{o}} \left( T_{1} \right) \sim 1.2 \times 10^{-19} \ J \ s^{-1} \ m^{-2} \ Hz^{-1} \ ster^{-1} \\
\sim 1.2 \times 10^{-16} \ erg \ s^{-1} \ cm^{-2} \ Hz^{-1} \ ster^{-1} \\
B_{\nu_{o}} \left( T_{2} \right) \sim 5.6 \times 10^{-28} \ J \ s^{-1} \ m^{-2} \ Hz^{-1} \ ster^{-1} \\
\sim 5.6 \times 10^{-25} \ erg \ s^{-1} \ cm^{-2} \ Hz^{-1} \ ster^{-1}
$$
Below is a table of values illustrating the changes I mention.
Temperature [K] | Ratio | Ratio
| [10^(11) Hz] | [10^(13) Hz]
------------------------------------------------------
1.0000000 | 2.2 x 10^(8) | 5.1 x 10^(833)
10.000000 | 16.2 | 2.3 x 10^(83)
100.00000 | 5.5 | 2.2 x 10^(8)
1000.0000 | 5.0 | 16.2
10000.000 | 5.0 | 5.5
100000.00 | 5.0 | 5.0
If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?
All of the above are intensities of light emitted by the blackbody source (i.e., the sun in this case). The power per unit area of radiation received at Earth's orbit scales as $\propto R_{s2a}^{-2}$, where $R_{s2a}$ is the distance from the source to the absorbing body. To calculate the total energy absorbed per unit time, one would need the following:
$$
\begin{align}
E_{ss} & = 4 \ \pi \ r_{s}^{2} \ \sigma \ T_{s}^{4} \tag{5a} \\
E_{sa} & = \frac{ E_{ss} }{ 4 \ \pi \ R_{s2a}^{2} } \tag{5b} \\
E_{aa} & = 4 \ \pi \ r_{a}^{2} \ \sigma \ T_{a}^{4} \tag{5c} \\
E_{as} & = \pi \ r_{a}^{2} \ E_{sa} \tag{5d}
\end{align}
$$
where $E_{ss}$ is the total energy emitted by source at the source surface, $E_{sa}$ is the total energy emitted by source at the absorbing bodies orbit, $E_{aa}$ is the total energy emitted by absorbing body at the absorbing body surface, $E_{as}$ is the total energy received over disk of absorbing body from the source, $r_{s(a)}$ is the radius of the source(absorbing body), and $T_{s(a)}$ is the blackbody temperature of the source(absorbing body).
To determine the change in the blackbody temperature of the the absorbing body, ignoring albedo effects, etc., one can just equate $E_{aa} = E_{as}$, or the energy absorbed equals the energy radiated (i.e., equilibrium).
References
- G.B. Rybicki and A.P. Lightman, Radiative Processes in Astrophysics, First Edition, John Wiley & Sons, Inc., New York, NY, 1979.