Mass of $H_2SO_4$ in $100\;ml$ (of $93 % \; H_2SO_4 $ )=$93 \;g$
So mass of $H_2SO_4$ in $1000\;ml(IL)$ solution.
$\qquad= \large\frac{93}{100}$$ \times 1000$
$\qquad= 930\;g$
Mass of $1000\; ml \; H_2SO_4$ Solution =volume $\times$ density
$\qquad= 1000 \times 1.84=1840\;g$
Hence mass of water in $1000\;ml\;H_2SO_4$ solution
$\qquad=(1840-930)g=910\;g$
Moles of $H_2SO_4=\large\frac{930}{98}$
Moles of $H_2SO_4$ in $1\;kg$ of water =$\large\frac{930}{98} \times \large\frac{1000}{910}$
$\qquad=10.43\;moles$
Hence molality of solution $=10.43$
Hence c is the correct answer.