The $\left[{\mathrm{OH}}^{-}\right]$of $0.005 {\mathrm{MH}}_2{\mathrm{SO}}_4$ is

It is given to find the $\left[{\mathrm{OH}}^{-}\right]$ of $0.005 {\mathrm{MH}}_2{\mathrm{SO}}_4$
${\mathrm{H}}_2{\mathrm{SO}}_4$ is a strong diprotic acid. The equation for ionization is
${\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}$
Each molecule of acid gives 2 molecules of ${\mathrm{H}}^{+}$.
The concentration of ${\mathrm{H}}^{+}$ion is $2\times 0.005=0.010\mathrm{M}$
$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log [0.01]=2.$
$\mathrm{pH}=2$
Hence the correct answer (A) $2\times {10}^{-12}\mathrm{M}$.