Calculate the molarity of solutions 'An' and 'Bn' based on the provided instructions.

The relative density (R.D.) of concentrated H₂SO₄ is 1.84. This means its density is $1.84 \text{ g/cm}^3$.
The volume of concentrated H₂SO₄ per $\text{dm}^3$ of solution is $5.60 \text{ cm}^3$.

$$\text{Mass of concentrated H}_2\text{SO}_4 = \text{Density} \times \text{Volume}$$ $$\text{Mass of concentrated H}_2\text{SO}_4 = 1.84 \text{ g/cm}^3 \times 5.60 \text{ cm}^3 = 10.304 \text{ g}$$

The purity of the concentrated H₂SO₄ is $98\%$.

$$\text{Mass of pure H}_2\text{SO}_4 = \text{Mass of concentrated H}_2\text{SO}_4 \times \text{Purity}$$ $$\text{Mass of pure H}_2\text{SO}_4 = 10.304 \text{ g} \times 0.98 = 10.09792 \text{ g}$$

The molar mass of H₂SO₄ is $98.0 \text{ g/mol}$.

$$\text{Moles of H}_2\text{SO}_4 = \frac{\text{Mass of pure H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4}$$ $$\text{Moles of H}_2\text{SO}_4 = \frac{10.09792 \text{ g}}{98.0 \text{ g/mol}} = 0.10304 \text{ mol}$$

The problem states this amount of H₂SO₄ is present per $\text{dm}^3$ of solution.

$$\text{Molarity of solution 'An'} = \frac{\text{Moles of H}_2\text{SO}_4}{\text{Volume of solution in dm}^3}$$ $$\text{Molarity of solution 'An'} = \frac{0.10304 \text{ mol}}{1 \text{ dm}^3} = 0.10304 \text{ M}$$

Rounding to three significant figures:

$$\boxed{\text{Molarity of solution 'An'} = \textbf{0.103 M}}$$

The mass of NaOH per $\text{dm}^3$ of solution is $4.0 \text{ g}$.
The molar mass of NaOH is $40.0 \text{ g/mol}$.

$$\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}}$$ $$\text{Moles of NaOH} = \frac{4.0 \text{ g}}{40.0 \text{ g/mol}} = 0.10 \text{ mol}$$

The problem states this amount of NaOH is present per $\text{dm}^3$ of solution.

$$\text{Molarity of solution 'Bn'} = \frac{\text{Moles of NaOH}}{\text{Volume of solution in dm}^3}$$ $$\text{Molarity of solution 'Bn'} = \frac{0.10 \text{ mol}}{1 \text{ dm}^3} = 0.10 \text{ M}$$

Rounding to two significant figures:

$$\boxed{\text{Molarity of solution 'Bn'} = \textbf{0.10 M}}$$