${{n}_{{{C}_{6}}{{H}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{6}}}}}=\frac{15,6}{78}=0,2\left( mol \right)$

$\Rightarrow {{n}_{{{C}_{6}}{{H}_{6pu}}}}=90%.{{n}_{{{C}_{6}}{{H}_{6}}}}=90%.0,2=0,18\left( mol \right)$

${{C}_{6}}{{H}_{6}}+B{{r}_{2}}\underrightarrow{Fe,{{t}^{0}}} &  &  & {{C}_{6}}{{H}_{5}}Br+HBr$

  0,18         $\to $     0,18                      (mol)

Theo phương trình ta có: ${{n}_{{{C}_{6}}{{H}_{5}}Br}}={{n}_{{{C}_{6}}{{H}_{6pu}}}}=0,18\left( mol \right)$

$\Rightarrow {{m}_{{{C}_{6}}{{H}_{5}}Br}}={{n}_{{{C}_{6}}{{H}_{5}}Br}}.{{M}_{{{C}_{6}}{{H}_{5}}Br}}=0,18.157=28,26\left( g \right)$

$\Rightarrow $Chọn đáp án B