Calculate the moles of sodium hydroxide (NaOH) used.

The volume of $\text{NaOH}$ is $25.0 \text{ cm}^3$, which needs to be converted to $\text{dm}^3$.

$$\text{Volume} = 25.0 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.0250 \text{ dm}^3$$

Now, calculate the moles of $\text{NaOH}$:

$$\text{Moles of NaOH} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of NaOH} = 1.60 \text{ mol/dm}^3 \times 0.0250 \text{ dm}^3 = 0.0400 \text{ mol}$$

The balanced equation is:

$$2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} + 8\text{H}_2\text{O(l)} \rightarrow \text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O(s)}$$

From the equation, 2 moles of $\text{NaOH}$ produce 1 mole of $\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}$.

$$\text{Moles of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = \frac{1}{2} \times \text{Moles of NaOH}$$ $$\text{Moles of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = \frac{1}{2} \times 0.0400 \text{ mol} = 0.0200 \text{ mol}$$

Using atomic masses: $\text{Na} = 22.99 \text{ g/mol}$, $\text{S} = 32.07 \text{ g/mol}$, $\text{O} = 16.00 \text{ g/mol}$, $\text{H} = 1.008 \text{ g/mol}$.

$$\text{Molar mass of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = (2 \times 22.99) + 32.07 + (4 \times 16.00) + 10 \times ((2 \times 1.008) + 16.00)$$ $$= 45.98 + 32.07 + 64.00 + 10 \times (2.016 + 16.00)$$ $$= 142.05 + 10 \times 18.016$$ $$= 142.05 + 180.16$$ $$= 322.21 \text{ g/mol}$$

$$\text{Mass} = \text{Moles} \times \text{Molar mass}$$ $$\text{Mass of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = 0.0200 \text{ mol} \times 322.21 \text{ g/mol}$$ $$\text{Mass of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = 6.4442 \text{ g}$$

Rounding to three significant figures (based on the given data $25.0 \text{ cm}^3$ and $1.60 \text{ mol/dm}^3$):

$$\text{Mass of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = \boxed{\text{6.44 g}}$$