The formula shows that Ni:O =0.98:1.00 or 98:100
thus if there are 100 O atoms , then Ni atoms =98
charge on 100 O2- ions =-200
suppose Ni atoms as Ni2+=x
then Ni atoms as Ni3+=98-x
Total charge on Ni2+ and Ni3+
=(+2)x +(+3)(98-x)
=294 - x
As metal oxide is neutral , total charge on cations =total charge on anions
294-x =200
x=94
% of Ni as Ni2+=94*100/98 = 95.9
% of Ni as Ni3+=100 - 95.9 =4.1