\(n_{Br_2}=n_{C_2H_4}=0.1\cdot2=0.2\left(mol\right)\)
\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(\Rightarrow n_{CH_4}=0.3-0.2=0.1\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\%CH_4=\dfrac{0.1}{0.3}\cdot100\%=33.33\%\)
\(\%C_2H_4=66.67\%\)