Partial pressure of $O_2$ , $P'_{O_2} = P_m\times mole\; fraction$
Since $P_m$ = 1 atm (given NTP condition)
$P'_{O_2} = 1\times mole\;fraction$-----(i)
For mixture
$PV = \large\frac{w}{m}RT$
$m = \large\frac{w}{VP}RT$
$m=\large\frac{1.3\times0.0821\times273}{1}(\therefore \large\frac{w}{V}=1.3g/l)$
$\therefore \text{mol.wt of mixture} = 29.137$
$If \;n_1 \;and\; n_2\; are\; mole\; of\; O_2\; and\; N_2\; respectively$
$Now , \large\frac{32\times n_1+28\times n_2}{(n_1+n_2)}=29.137$
$\large\frac{28n_1+28n_2}{n_1+n_2} + \large\frac{4n_1}{n_1+n_2} = 29.137$
$\therefore \large\frac{n_1}{n_1+n_2} = \large\frac{29.137-28}{4} = 0.28$
$\therefore Mole\;fraction\;of\;O_2 = 0.28$
$\therefore By\; Eq (1) , P'_{O_2} = 0.28 \;atm$