22.414mL of volume at STP is occupied by =0.078g of gaseous hydrocarbon
$\therefore$ 22400mL volume at STP is occupied by =$\large\frac{0.078g}{22.414mL}$$\times 22400mL=78$g gaseous hydrocarbon.
$\therefore$ Molar mass of hydrocarbon=78gmol$^{-1}$
Empirical formula mass of hydrocarbon $CH=12+1=13$
$\therefore n=\large\frac{\text{Molecular mass}}{\text{E.F.mass}}=\frac{78}{13}$$=6$
$\therefore$ Molecular formula of hydrocarbon =$(CH)_6=C_6H_6$
Hence (b) is the correct answer.