The number of equivalents of $\mathrm{HCl}$ in  $200\mathrm{ml}$ of  $0.1\mathrm{N} \mathrm{HCl}$ solution is

 We know that Normality$\left(\mathrm{N}\right)=\frac{\mathrm{No}. \mathrm{of} \mathrm{gram} \mathrm{equivalents}\left(n\right)}{\mathrm{Volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{litres})}$

Normality $=0.1\mathrm{N}$(Given)

Volume of solution $=200\mathrm{ml}=\frac{200}{1000}=0.2 \mathrm{l}$

Therefore,

According to formula

$0.1=\frac{\mathrm{n}}{0.2}$

$\mathrm{n}=0.1\times 0.2$ 

$\mathrm{n}=0.02$

Hence, option(c) is correct.