P = $\large\frac{800 \text{ mm Hg}}{760}=\large\frac{800}{760}atm$
T = 273+100 = 300K
Let the density be d for $CO_2$
From Ideal gas equation, for $CO_2$ PV = $\large\frac{w}{m}RT$
P = $\large\frac {d}{m} RT$
(Since d = $\large\frac{w}{v}$)
$\large\frac{800}{760} = \large\frac{d}{44}\times0.0821\times373$
$\therefore d = 1.5124g/l$
Hence answer is (a)