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.