This is a confusing question because, while solubilities can be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units:
$$ \rho = \frac{cRT}{p}$$
In this online data page, for instance , in some columns the solubility is reported in mL/L, converted to this unit using throughout a pressure of $\pu{1 atm}$ (the source of the data cannot be verified), even as the partial pressure of oxygen $p_{O_2}$ is increased above $\pu{1 atm}$.
In the OP the volume refers presumably to an equivalent volume of oxygen gas at the (partial) pressure of the gas above the liquid.
Using the numbers from the OP, assuming the gas is ideal, then
$ c=\frac{101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 1.1 \times 10^{-3}M}$
when $p = \pu{1 atm}$.
On the other hand, if $p = \pu{4 atm}$
$ c=\frac{4\times101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 4.6\times 10^{-3} M}$
so the solubility is the same ($\pu{28 mL/L}$) if described in terms of volume at the given pressure, but $\times 4$ greater when regarded as a molar concentration.
Note by the way that according to a number of sources the solubility at $\pu{25 ^\circ C}$ is $\pu{258 \mu M}$ (~$\pu{8.2 mm Hg}$) at $p_{O_2}=\pu{1 atm}$, and $\pu{1.0 mM}$ at $p_{O_2}=\pu{4 atm}$.
