**Given:**
– O₂ concentration in water required for fishes = 3.8 mg/L
– Solubility of O₂ in water = 2.2 × 10⁻³ mol/L
– Pressure = 1 atm

**To find:** Partial pressure of O₂ above the water needed for the survival of fish.

**Formula:** \(S = K_H \cdot P\)

**Calculation:**
– Pressure = 1 atm = 1.013 bar

Now, using the formula and rearranging,

\[K_H = \frac{S}{P} = \frac{2.2 \times 10^{-3} \, \text{mol/L}}{1.013 \, \text{bar}} = 2.17 \times 10^{-3} \, \text{mol L}^{-1} \, \text{bar}^{-1}\]

O₂ concentration in water required for fishes,

\[= 3.8 \, \text{mg/L} = \frac{3.8 \times 10^{-3} \, \text{g/L}}{32 \, \text{g/mol}} = 1.19 \times 10^{-4} \, \text{mol L}^{-1}\]

Now, using the formula and rearranging,

\[P = \frac{S}{K_H} = \frac{1.19 \times 10^{-4} \, \text{mol L}^{-1}}{2.17 \times 10^{-3} \, \text{mol L}^{-1} \, \text{bar}^{-1}} = 0.0548 \, \text{bar}\]

The partial pressure of O₂ above the water needed for the survival of fish is 0.0548 bar.