At STP 22400ml of $NO_2$=46g of $NO_2$
$112.0$ml of $NO_2=\large\frac{112.0mL\times 46.0g}{22400mL}$$=0.23$g
$V_{NO_2}=\large\frac{Mass}{Density}=\frac{0.23}{1.15gmL^{-1}}$$=0.20mL$
Number of molecules =$\large\frac{0.23}{46}$$\times 6.02\times 10^{23}$
$\Rightarrow 3.01\times 10^{21}$
Hence (b) is the correct answer.