$n_{H^+(A)}=2n_{H_2SO_4}=0,5V_1.2=V_1(mol)$
$n_{OH^-(B)}=n_{NaOH}=0,6V_2(mol)$
Sau phản ứng, $pH<7$ nên dư $H^+$
$n_{H^+\text{dư}}=V_1-0,6V_2(mol)$
$[H^+]=10^{-1}=0,1M$
$\to V_1-0,6V_2=0,1(V_1+V_2)$
$\to 0,9V_1=0,7V_2$
$\to \dfrac{V_1}{V_2}=\dfrac{7}{9}$