98 Ni atoms are associated with 100 O-atom. Out of 98 Ni atoms, suppose Ni present as $Ni^{2+} = x$

Then Ni present as $Ni^{3+} = 98 – x$

Total charge on $xNi^{2+}$ and $(98 – x)Ni^{3+}$ should be equal to charge n $100O^{2−}$. Hence

$x \times 2 + (98 – x) \times 3 = 100 \times 2$

$2x + 294 – 3x = 200$

x = 94

Fraction of Ni present as $Ni^{2+ }=\large\frac{94}{98}$$\times 100=96\%$

Fraction of Ni present as $Ni^{3+ }=\large\frac{4}{98}$$\times 100=4\%$

Hence (A) is the correct answer.