How much water evaporates in a vacuum before freezing?
You will need to know:


the heat capacity of water $C_p$
the latent heat of fusion of ice $L_f$
the latent heat of vaporization of water $L_v$


First determine how much heat must be lost to change the temperature from 20 to 0 using the heat capacity.

Next determine how much heat must be lost to change from 0 degree water to 0 degree ice using the latent heat of fusion.  

Finally, add the two heats from above together to get a total heat loss, and use the heat of vaporization of water to calculate the amount of water evaporated that corresponds to this total heat loss.

Note: The above involves an approximation that the amount of water evaporated is relatively small.  In reality, once an infinitesimal portion of water evaporates, the amount remaining that needs to cool will be infinitesimally less than one kg.  If you are expected to know and use intergral calculus for your class, you may need to take into consideration how the mass is varying thoughout the process.  

To calculate more accurately, break the problem is into two parts.  

First, a mass m1 evaporates, to cool water from 20 degres to 0 degrees.

Let $m$ be the varible liquild mass during the process and temperature be $t$

$mC_pdt=L_vdm$

$C_p\int_{20}^{0}dt = L_v \int_{1}^{1-m1}(dm/m)$

The remaining mass is 1kg-m1

Second, a mass m2 evaporates, to freeze the remaining water.

$L_v m2 = L_f (1kg - m1 - m2)$