You can take the decrease in volume of the solid into account by calculated how much mass has been lost, and using the density of the ice to determine how much volume was lost. In a container large enough to get substantial sublimation, I think you will find that the change in volume of the solid is small enough to be negligible, but it doesn't hurt to do the calculation.
I'm not sure what you mean by:
How would I calculate equilibrium vapour pressure of water at each temperature if it is not given?
The table at the bottom of the page gives the formulas it used for calculation:
Formulas
The table values for −100 °C to 100 °C were computed by the following formulas, where T is in kelvins and vapor pressures, Pw and Pi, are in pascals.
Over liquid water
loge(Pw) = –6094.4642 T−1 + 21.1249952 – 2.724552×10−2 T + 1.6853396×10−5 T2 + 2.4575506 loge(T)
For temperature range: 173.15 K to 373.15 K or equivalently −100 °C to 100 °C
Over ice
loge(Pi) = –5504.4088 T−1 – 3.5704628 – 1.7337458×10−2 T + 6.5204209×10−6 T2 + 6.1295027 loge(T)
For temperature range: 173.15 K to 273.15 K or equivalently −100 °C to 0 °C
Are you asking how the equations were derived?