The oxygen dissolved in water exerts a partial pressure of 20kPa in...

$P = {K}_{H} \cdot x$

or, $20 \times {10}^{3} = \left(8 \times {10}^{4} \times {10}^{3}\right) \times \dfrac{{n}_{{O}_{2}}}{{n}_{{O}_{2}} + {n}_{\text{water}}}$

or, $\dfrac{1}{4000} = \dfrac{{n}_{{O}_{2}}}{{n}_{{O}_{2}} + {n}_{\text{water}}} = \dfrac{{n}_{{O}_{2}}}{{n}_{water}}$

Means $1$ mole water $\left(= 18gm = 18ml\right)$ dissolves

$\dfrac{1}{4000}$ moles ${O}_{2} .$ Hence, molar solubility

$= \dfrac{\left(\dfrac{1}{4000}\right)}{18} \times 1000 = \dfrac{1}{72}mol{dm}^{- 3}$

$= 1388 . 89 \times {10}^{- 5}mol{dm}^{- 3} \approx 1389 \times {10}^{- 5}mol{dm}^{- 3}$