
20.21*10^-4 mg/L is the concentration.
Explanation:
Given: Â
Pressure=1 atm.
Temperature= Â 0 degrees
Volume percent of air is 21%
Henry's Law constant K = 2.28 x 10^-3 mole/L-atm
The partial pressure of oxygen is 0.21 atm.
By Henry's law:
Concentration= K X Partial pressure
            =  2.28*10^-3 *0.21
             =  4.79*10^-4 moles/litre
Since, at STP 1 mole of oxygen occupies volume of 22.4L
concentration =mass/volume
   mass          =  4.79*10^-4-4*22.4
                 = 20.21*10^-4 mg
  20.21*10^-4 mg/L is the concentration.