In the disproportionate reaction

$3{\mathrm{HClO}}_3⟶{\mathrm{HClO}}_4+{\mathrm{Cl}}_2+2{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

the equivalent mass of the oxidising agent is: (molar mass) of HCIO₃ = 84.45)

We know that, equivalent mass = Molar mass/oxidation number change.

Oxidation no. of Cl in HCIO₃ .$\begin{array}{r}\text{ O.N. of }\mathrm{H}+\mathrm{O}.\mathrm{N}.\text{ of }\mathrm{Cl}+3\times \mathrm{O}.\mathrm{N}.\text{ of }\mathrm{O}=0;+1+\mathrm{x}+3(-2)+0;\mathrm{x}=+5\end{array}$

O.N. of Cl in Cl₂ = 0. So, O.N. change= 5- 0 = 5 Equivalent mass = Molar mass/O.N. change = 84.45/5 i.e., 16.89
So, the correct answer is (1).