The $P^H$  of $NaOH$  solution is 12. Then the $P^H$  of $H_{2}S{O}_4$  solution having same molar concentration of $NaOH$  solution is

$\left[NaOH\right]={10}^{-2}M$

$\left[H_{2}S{O}_4\right]={10}^{-2}M$

$\left[H^{+}\right]=2\times {10}^{-2}M$

$P^H=-\log \left(2\times {10}^{-2}\right)$

        $=2-\log 2$

        $=2-0.301$

        $=1.699\approx 1.7$