Calculating 30% less solar energy. Provided the math is right, would there have been liquid water on Earth? The Sun is  estimated   to  have    had a   solar   luminosity  about   30% lower   when    the solar   system  was formed. 

Now if I assumed that Earth's albedo was the same as it is today, what would have been the effective radiating temperature of the Earth at that time.

My solution:
T_e = S/4*(1 - A)

I know that Earth's current albedo is A = 0.3.

Hence,

T_e = S/4*(1 - 0.3)

T_e = ( (0.7)*(1370) ) / (4 (5.67*10^-8))  * (1-0.3) = 233K

233K &lt;- would have been Earth's effective radiating temperature

Now if one assumes the magnitude of the greenhouse effect was the same then (now it is 33k) Earth's temperature would be:

233K + 33K = 266K

Assuming the math is right between those two would there have been liquid water on Earth? If there would be, how does one resolve this paradox? that is, given that albedo was the same, what element of the climate system could explain this?

Thanks
Quick and simple answer: A greater greenhouse effect.

There is an equilibrium in place. CO2 weathering rates depend on temperature, so at lower temperatures, we see CO2 levels rise, warming the planet. At higher temperatures, higher weathering rates mean more CO 2 drawdown, so the planet cools.

This is a very long term process. But this CO2 thermometer effect is believed to be the reason why we have had liquid oceans for all of known earth history.

As far as the black body effect goes; this is correct - from the viewpoint of an observer at distance, Earth would 'appear' colder. However, this is because the emission height changes (see https://andthentheresphysics.wordpress.com/2014/03/05/effective-emission-height/ ) for more of an explanation. Think of it as 'the height at which the atmosphere becomes IR-opaque' (although this is a simplification); the temperature then follows the lapse rate down to the ground. So raising the emission height raises the surface temperature. 

(NB late at night, add refs later..)