By definition, 3.6 M or `3.6 xx 98 = 352.8 g` of `H_(2)SO_(4)` are present in 1L of the solution

Now, 29g of `H_(2)SO_(4)` are present in the solution = 100 g

352.8 g of `H_(2)SO_(4)` are present in the solution `= ((100g)xx(352.8))/((29g))=1216g`

Density of solution `= ("Mass of solution")/("Volume of solution")=((1216g))/((1000mL))=1.22"g mL"^(-1)`