${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{2}}}}}{22,4}=\frac{6,72}{22,4}=0,3\left( mol \right)$

${{C}_{2}}{{H}_{2}}+2B{{r}_{2}}\to {{C}_{2}}{{H}_{2}}B{{r}_{4}}$

 0,3 $\to $  0,6                              (mol)

Theo phương trình ta có:

${{n}_{B{{r}_{2}}}}=2.{{n}_{{{C}_{2}}{{H}_{2}}}}=2.0,3=0,6\left( mol \right)$

$\Rightarrow {{V}_{B{{r}_{2}}}}=\frac{{{n}_{B{{r}_{2}}}}}{{{C}_{{{M}_{B{{r}_{2}}}}}}}=\frac{0,6}{0,2}=3\left( l \right)$

$\Rightarrow $Chọn đáp án D