10 ml of 1 M ${\mathrm{H}}_2{\mathrm{SO}}_4$ will completely neutralize

${\mathrm{H}}_2{\mathrm{SO}}_4 + 2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 3{\mathrm{H}}_3{\mathrm{O}}^{+} + {\mathrm{SO}}_4^{- -}$.                                                            

$\text{NaOH} \rightleftharpoons {\mathrm{Na}}^{+}+ {\mathrm{OH}}^{-}$

 $$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{equivalents} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_{4 } = \mathrm{N} {\mathrm{V}}_{\mathrm{lit}} \\ =2\times \frac{10}{1000} [\mathrm{N}=\mathrm{M}\times \mathrm{Basicity}=1\times 2] \\ \mathrm{Number} \mathrm{of} \mathrm{Equivalents} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4=\mathrm{Number} \mathrm{of} \mathrm{Equivalents} \mathrm{of} \mathrm{NaOH} \end{gathered}$$ ,  

In Option(2) , $\mathrm{n}=\frac{\mathrm{M}\times \mathrm{V}\left(\mathrm{ml}\right)}{1000}=\frac{2\times 10}{1000}=2\times {10}^{-2} \mathrm{moles}$

Hence option(2) is the answer