The correct option is C (i) < (ii) < (iii)
Sodium metal replaces hydrogen of −OH bond. It means higher the replacability of hydrogen of -OH or acidity of the moleucle, higher will be the reactivity with sodium metal.
More the number of methyl groups (electrons releasing) lesser will be the acidity. Hence the reactivity towards sodium metal will be tertiary alcohol < secondary alcohol < primary alcohol
Hence, the correct answer is option (c).