Assuming complete ionisation, the pH of 0.1 M HCI is 1. The molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with the same pH is

$\left[{\mathrm{H}}^{+}\right]\text{in }0.1\mathrm{MHCl}=0.1$

$\text{ For same }\mathrm{pH},\left[{\mathrm{H}}^{+}\right]\text{in }{\mathrm{H}}_2{\mathrm{SO}}_4=0.1\mathrm{M}$

$\text{ One mole }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ gives }2\text{ moles of of }{\mathrm{H}}^{+}\text{ions }$

${\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}$

$\therefore 0.1\mathrm{M}\left[{\mathrm{H}}^{+}\right]\text{will be given by }$

$=\frac{0.1}{2}=0.05{\mathrm{MH}}_2{\mathrm{SO}}_4$