This would be a standard question for someone starting in chemistry (and thus there is no reference as it's 'obvious' to someone skilled in the art), so obviously you are not a chemist, but your information is incomplete to answer it  You do not state the form of the precursor 'cerium nitrate'.  You finish with CeO2 (cerium in the +4 oxidation state).  The most common nitrate of cerium is the hexahydrate [Ce(NO3)3.6H2O] and this is Ce in the +3 oxidation state - other nitrates are, of course, known - hydrated and unhydrated and Ce in various oxidation states.  However, this can be taken as a pertinent example.  The molecular mass of the Ce (+3) nitrate hexahydrate is 434.22. The molecular mass of CeO2 is 172.115.  So 434.22 g of the hexahydrate will lead to 172.115 g of CeO2 (100% yield). So if you start with 1 mg of the nitrate you'll end up with 172.115/434.22 mg of the oxide. This is around 0.4 mg (0.396 is more pedantic).  So the chemistry of your starting material is crucial.  You ask 'how much?'  I think in terms of nanoparticles then the question should be 'how many?'  You would need then to calculate how many NP's of say 50 nm would be produced from (say) 0.4 mg of the oxide.  This should be an easy question for you to answer with a knowledge of the density of the CeO2 and some assumption about the finished shape of particle.

Sir, Thanks for giving your valuable time to answer my question. Yes, i meant Ce(No3)3.6H20. Your answer is very helpful. Actually there are multiple references where CeO2.NP were produced from the mentioned salt. The conversion of Salt to CeO2 and finally to Narrow ranged size NP (~50nm or less) would surely lead to a  final yield which i guess may or may not have significant variation from the estimated one. A mention of the actual yield obtained by researchers in their reports may have been more helpful.    

In a reduction (or nitrate decomposition) reaction, one would expect all the nitrate to be converted to oxide. If all the material remains in suspension with no loss due to precipitation, sedimentation etc then the yield is effectively 100%. Carry out a Stokes' Law calculation. CeO2 is dense (6.7 g/cm3 if I remember correctly) and would have to be small (< 50 nm?) to stay in suspension for a significant amount of time (assumed to be > 10 days).

Sir, thanks for quick reply. Your response is helpful!