At Standard Temperature and Pressure (STP), $1\text{ mole}$ of an ideal gas occupies a volume of $22.4\text{ L}$. Molar mass of $\text{CCl}_4 = 12 + 4(35.5) = 154\text{ g/mol}$. Density is the mass per unit volume. Therefore, the density of $\text{CCl}_4$ vapour at STP is: $\text{Density} = \frac{\text{Molar Mass}}{\text{Molar Volume}} = \frac{154\text{ g/mol}}{22.4\text{ L/mol}} = 6.875\text{ g/L}$. The calculated density is $6.875\text{ g/L}$, which is nearest to the option $6.84$.