Calculate the molarity and molality of 93% H2SO4 (weight/volume).
Density of solution is 1.84g per ml

OK, assuming it is 93% by mass then we should work with 100 g of solution

a 93% by mass solution will have 93 g of H2SO4 and 7 g of H2O

moles H2SO4 = mass / molar mass = 93 g / 98.086 g/mol = 0.948148 moles
mass H2O = 7 g = 0.007 kg

molality = moles solute / kg solvent
= 0.948148 mol / 0.007 kg
= 135 m
~ 140 m (2 sig figs)

Now
total volume of 100 g of solution = mass / density
= 100 g / 1.84 g/ml
= 54.35 ml
= 0.05435 L

molarity = moles solute / litres solution
= 0.98148 mol / 0.05435 L
= 18 M