Calculate the mass of MgSO4 ext· 7H2O required for a 0.25 L solution with 0.1 M molarity.

$$\text{Moles of MgSO}_4 = \text{Molarity} \times \text{Volume (L)}$$ $$\text{Moles of MgSO}_4 = 0.1 \text{ mol/L} \times 0.250 \text{ L} = 0.025 \text{ mol}$$

Atomic masses: Mg = 24, S = 32, O = 16, H = 1.
Molar mass of MgSO$_4$:

$$\text{Mg} = 1 \times 24 = 24$$ $$\text{S} = 1 \times 32 = 32$$ $$\text{O} = 4 \times 16 = 64$$ $$\text{Molar mass of MgSO}_4 = 24 + 32 + 64 = 120 \text{ g/mol}$$

Molar mass of 7H$_2$O:

$$\text{H}_2\text{O} = (2 \times 1) + 16 = 18 \text{ g/mol}$$ $$7\text{H}_2\text{O} = 7 \times 18 = 126 \text{ g/mol}$$

Molar mass of MgSO$_4 \cdot 7$H$_2$O:

$$\text{Molar mass} = 120 \text{ g/mol} + 126 \text{ g/mol} = 246 \text{ g/mol}$$

Since 1 mole of MgSO$_4 \cdot 7$H$_2$O contains 1 mole of MgSO$_4$, the moles of MgSO$_4 \cdot 7$H$_2$O needed are the same as the moles of MgSO$_4$.

$$\text{Mass of MgSO}_4 \cdot 7\text{H}_2\text{O} = \text{Moles} \times \text{Molar mass}$$ $$\text{Mass of MgSO}_4 \cdot 7\text{H}_2\text{O} = 0.025 \text{ mol} \times 246 \text{ g/mol} = 6.15 \text{ g}$$

The mass of MgSO$_4 \cdot 7$H$_2$O required is $\boxed{\text{6.15 g}}$.