93% $${H_2}S{O_4}$$ solution weight by volume means in 100 ml solution 93 gm $${H_2}S{O_4}$$ present.

Density of solution = 1.84 g/mL

$$\therefore$$ Weight of solution = 100 $$\times$$ 1.84 = 184 gm

$$\therefore$$ Weight of solvent $${H_2}O$$ = 184 $$-$$ 93 = 91 gm

Now,

$$\mathrm{Molality ={{Moles\,of\,solute} \over {Weight\,of\,solvent\,in\,Kg}}}$$

$$ = {{{{93} \over {98}}} \over {{{91} \over {1000}}}} = 10.42$$