Ignoring meteorological factors and any dust or satellites, this is still an incomplete problem. You would also need to know the rotation axis. For example, Uranus rotates completely on its side (i.e. it rotates at 97.77 degrees, while earth rotates at 23 degrees, 26 minutes and 21.4119 seconds). Such things become important for factors like the Arctic Circle. You are also missing the distance between the planet and the star. That becomes important when you consider that Pluto is colder than Mercury. This is needed for the formulation of the sterdian.
In your question, you said to ignore atmospheric effects. However, the atmosphere also acts as a secondary source of radiation (the infamous greenhouse effect). Even if you neglect eddies, advection, etc. you also need to consider that the atmosphere can absorb radiation and emit it back to the surface, otherwise you would find the average temperature of the surface to be the same as the average temperature of the earth.
Edit: I found a decent approximation (approximation being the key word).
Let $Q_S$ is the incoming solar radiation flux
$$Q_s= S(1-\alpha)(\frac{\bar{d}}{d})^2cos(\zeta)\tau_s $$
Where $S$ is the solar constant, $\alpha$ is the albedo, $d$ is the distance of the earth from the sun, $\bar{d}$ is the mean distance of the earth from the sun and $\tau_s$ is the transmissivity of the atmosphere (including any clouds above).
The last factor, $cos(\zeta)$ can be computed by$$cos(\zeta)=sin(\psi)sin(\delta)+cos(\psi)cos(\delta)cos(h)$$ where $$\delta=23.45\times\frac{\pi}{180}cos(\frac{2\pi\times(d-d_{solst})}{d_{year}})$$
Where $d$ is the Julian day, $d_{solst}$ is the julian day of the solstice (173) and $d_{year}$ is the number of days in the year (365.25)
Additionally, $h$ is the local hour, as defined by $$h=\frac{(t_{UTC}-12)\times\pi}{12}+\frac{\lambda\pi}{180}$$ where $\lambda$ is the longitude and $t_{UTC}$ is the time in UTC (in hours).
