Answer
5.23 g/L
Work Step by Step
Recall: $d=\frac{PM}{RT}$ where $d$ is the density, $P$ is the pressure, $M$ is the molecular weight, $R$ is the universal gas constant and $T$ the temperature in K.
Given/Known: $P=1.00\,atm, M=188\,g/mol, $
$R=0.0821\,L\,atm\,mol^{-1}K^{-1}$ and
$T=(165+273)K= 438\,K$
Substitute:
$d=\frac{1.00\,atm\times188\,g/mol}{(0.0821\,L\,atm\,mol^{-1}K^{-1})(438\,K)}$
$=5.23\,g/L$
