An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is

Mole fraction of solute ${\mathrm{X}}_2=0.2$. Therefore, mole fraction of solvent ${\mathrm{X}}_1=0.8$

Or $\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.2$ and $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.8$

Or $\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{0.2}{0.8}=\frac{1}{4}$

Now, if ${\mathrm{n}}_1$ (solvent moles) = 1000/78 = 12.8 moles 

${\mathrm{n}}_2=12.8/4=3.2$ moles. Therefore, 3.2 moles of the compound are present in one kg of solvent benzene and so molality = 3.2.

                                                                                   $\left[OR\right]$

$$\begin{gathered} \mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{solute} = \frac{ \mathrm{m}}{\mathrm{m}+{\displaystyle \frac{1000}{\mathrm{Molecular} \mathrm{weight} \mathrm{of} \mathrm{solvent}}}} \\ 0.2 = \frac{\mathrm{m}}{\mathrm{m}+{\displaystyle \frac{1000}{78}}} \\ \Rightarrow 0.2 =\frac{ \mathrm{m}}{\mathrm{m}+12.82} \\ \mathrm{m} = 0.2\mathrm{m}+ \left[12.82\times 0.2\right] \\ \mathrm{m} = 3.2 \end{gathered}$$