Answer
$2.2\space kg/m^{3}$
Work Step by Step
We can write,
$Density\space (\rho)=\frac{mass(m)}{volume(V)}=\frac{n(Mass\space per\space mole)}{V}-(1)$
Let's apply the ideal gas law $PV=nRT$ to the substance.
$PV=nRT=> n=\frac{PV}{RT}-(2)$
(2)=>(1),
$\rho=\frac{(\frac{PV}{RT})(mass\space per\space mole)}{V}=\frac{P(mass\space per\space mole)}{RT}$
Let's plug known values into this equation.
$\rho=\frac{2(1.013\times10^{5}Pa)(28\times10^{-3}g/mol)}{(8.31\space J/mol\space K)(310\space K)}=2.2\space kg/m^{3}$
Density of the substance = $2.2\space kg/m^{3}$