2.0 g of a bromo hydrocarbon (X) was subjected to Carius analysis,...

Mass of $AgBr$ formed = $3.36$ g
Molar mass of $AgBr = 108 + 80 = 188$ g mol$^{-1}$

Moles of $AgBr$ formed = $\dfrac{3.36}{188} \approx 0.01787$ mol
Since $1$ mole of $AgBr$ contains $1$ mole of $Br$, moles of $Br$ in the compound = $0.01787$ mol.
Mass of $Br = 0.01787 \times 80 = 1.43$ g

Percentage of $Br$ in compound $(X) = \dfrac{1.43}{2.0} \times 100 = 71.5\%$
Given percentage of $C = 26.7\%$
Percentage of $H = 100\% - (71.5\% + 26.7\%) = 1.8\%$

Now, we find the molar ratio of the elements in the compound:
Moles of $C = \dfrac{26.7}{12} = 2.225$
Moles of $H = \dfrac{1.8}{1} = 1.8$
Moles of $Br = \dfrac{71.5}{80} = 0.893$

Dividing by the smallest value ($0.893$) to get the simplest ratio:
$C : H : Br = \dfrac{2.225}{0.893} : \dfrac{1.8}{0.893} : \dfrac{0.893}{0.893}$
$C : H : Br \approx 2.5 : 2 : 1$

Multiplying by $2$ to obtain whole numbers, we get the ratio $5 : 4 : 2$.

The empirical formula of the compound is $C_5H_4Br_2$.
The total number of carbon atoms in the empirical formula is $5$.

Answer: $5$