The molarity of a  $9.8\%\left[\frac{\mathrm{W}}{\mathrm{W}}\right] {\mathrm{H}}_2{\mathrm{SO}}_4$ of density $1.1 \mathrm{g}/\mathrm{cc}$ is:

$9.8\%\left[\frac{\mathrm{W}}{\mathrm{W}}\right]$ of  ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution means $9.8 \mathrm{g}$ of solute in$100 \mathrm{g}$ of solution.

We know that, Density $=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

Therefore, Volume$=\frac{\mathrm{Mass}}{\mathrm{Density}}$

Then,

Density of  ${\mathrm{H}}_2{\mathrm{SO}}_4=1.1 \mathrm{g}/\mathrm{ml}$

We know that $1\mathrm{g}/\mathrm{cc}=1\mathrm{g}/\mathrm{ml}$

The volume of the solution $=\frac{\text{ Mass }}{\text{ Density }}=\frac{100}{1.1}=90.9\mathrm{ml}=\frac{90.9}{1000} \mathrm{l}$

Molarity $=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute} }{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{litres}}$

No. of moles of solute$\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=\frac{\mathrm{Given} \mathrm{Mass}}{\mathrm{Molar} \mathrm{Mass}}=\frac{9.8}{98}=0.1$

Then

Molarity $=\frac{0.1}{{\displaystyle \frac{90.9}{1000}}}=\frac{0.1\times 1000}{90.9}=1.1 \mathrm{M}$

Hence, option (d) is correct