The volume of 0.1 M strong dibasic acid sufficient to neutralize 1g...

For neutralisation;
$⇒ \text{n}\text{o} . \text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}{\text{H}}^{+} = \text{n}\text{o} . \text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{O}{\text{H}}^{-}$
$⇒ 2 \times {\text{M}}_{\text{a}}{\text{V}}_{\text{a}} = 0.04$
$⇒ 2 \times 0.1 \times {\text{V}}_{\text{a}} = 0.04$
$⇒ {\text{V}}_{\text{a}} = \dfrac{0.04}{0.2} = 0.2\text{L} = 200\text{mL}$