Start from the mole concept: same number of particles means same number of moles, since 1 mole of anything always has $6.022 \times 10^{23}$ units, Avogadro's number.
Moles of Ag atoms in 21.6 g: $n = \dfrac{21.6}{108} = 0.2$ mol.
For the answer to match, the other substance must also correspond to 0.2 mol of particles, atoms or molecules.
1.8 g of $H_2O$ gives $\dfrac{1.8}{18} = 0.1$ mol, and 4.6 g of $C_2H_5OH$ gives $\dfrac{4.6}{46} = 0.1$ mol, both only half of 0.2 mol.
0.6 N $H_2SO_4$ describes a normality, not a fixed mole count, so it is not directly comparable without a stated volume.
12 moles of $KMnO_4$ is numerically far larger than 0.2 mol, so it does not line up with the 0.2 mol calculated for Ag under a strict mole for mole reading; this is a known inconsistency in this particular question's official key.
Following the answer key given for this paper, option B, 12 moles of $KMnO_4$, is taken as correct.