Find the molarity and molality
All in all this isn't a well stated problem. The molarity can be solved, but the molality can't really be solved.
The wording &quot;with a 96% of mass&quot; is odd. I'm not sure if this is your interpretation of the problem or if that is what is literately written in the book. I'll interpret the phrase to mean that the sulfuric acid was 96% sulfuric acid by weight.
For both problems the grams of sulfuric acid are:
$$\mathrm{grams\ }\ce{H2SO4} = 10 \times 1.84 \times 0.96 = 17.66$$
$$\mathrm{moles\ }\ce{H2SO4} = \dfrac{10 \times 1.84 \times 0.96}{98.079} = 0.1801 $$
Molarity
So the molarity, $M$, of the solution will be:
$$\mathrm{Molarity} = \dfrac{10 \times 1.84 \times 0.96}{98.079 \times 0.50} = 0.360 $$
Molality(1)
The problem with the molality calculation is that volumes are not additive. I'll admit that for two or three significant figures the calculation sort of works.
So for the 500 ml of solution, you must assume that 10 ml is sulfuric acid, the other 490 ml of volume is water. Water is assumed to be 1.00 kg/liter, so to figure out how many moles of sulfuric acid per kilogram of solvent we'll need to multiply the 0.360 moles per liter of solution by a correction factor of 1/0.980 which gives a value of 0.368 for the molality.
Molality(2)
As a second way to calculate the molality let's divide moles of $\ce{H2SO4}$ by the kilograms of water.
From above we know that the 500 ml of solution contained 0.1801 moles of $\ce{H2SO4}$.
Again we have to assume that the volumes are additive so that 10 ml of concentrated sulfuric acid and 490 ml of water yield 500 ml of solution. Given no temperature we can just assume that water is 1.00 g/ml, so 490 ml of water has a mass of 490 grams or 0.490 kilograms.
$$\mathrm{Molality} = \dfrac{0.1801}{0.490} = 0.368 $$

To perhaps beat a dead horse here let's back calculate the volumes for one liter of a similar solution. Steffen's Chemistry Pages give that a 4% solution of sulfuric acid by weight has a density of 1.0250 g/ml, 41.00 g sulfuric acid per liter and a molarity of 0.418 at 20°C.
To get 41.00 grams of sulfuric acid we'd need
$$\mathrm{ml\ conc\ sulfuric\ acid} = \dfrac{41.00}{1.0250\times 0.96} = \pu{41.7 ml}$$
But since water is 1 g/ml and the solution is 96% water
$$\mathrm{ml water} = 1000 \times 0.96 \times 1.0250 = \pu{984 ml}$$
$$41.7 + 984 = 1025.7$$