Percent labeling of oleum $= 100 +$ Total mass of water required to completely convert oleum into${\mathrm{H}}_2{\mathrm{SO}}_4$

So, we need $18 \mathrm{g}$ of water for the reaction ${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4$

one mole of water is required, which in turn means that one mole of ${\mathrm{SO}}_3$ is present. Therefore, in a $100 \mathrm{g}$ sample, $80 \mathrm{g}$ is ${\mathrm{SO}}_3$_​.

Hence, the mass of ${\mathrm{H}}_2{\mathrm{SO}}_4= 20 \mathrm{g}$. 

Number of moles of ${\mathrm{SO}}_3=\frac{80}{80}=1$

Number of moles of${\mathrm{H}}_2{\mathrm{SO}}_4=\frac{20}{98}=0.20$

one mole of ${\mathrm{H}}_2{\mathrm{SO}}_4$​ requires two moles of $\mathrm{NaOH}$ for complete neutralization

one mole of ${\mathrm{SO}}_3$​ requires two moles of $\mathrm{NaOH}$ for complete neutralization

${\mathrm{SO}}_3+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}$

So, number of moles of $\mathrm{NaOH}$ required for neutralization of $100\mathrm{g}$ of $118\%$ oleum$=2\times$Number of ${\mathrm{SO}}_3$$=2\times$number of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$

$=2\times 1+2\times 0.2=2.4$ moles