Quantity of water in a redox problem
To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:

$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$

where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.

From the balanced redox reaction, which it seems you have established correctly:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -&gt; 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:

$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$

$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$

$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$

From (1):

$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$

Combining (2–4) into (5), and taking into account that

$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$

one can obtain the following expression for the volume of water to be added:

$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$

The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:

$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$