${{n}_{{{C}_{6}}{{H}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{6}}}}}=\frac{7,8}{78}=0,1\left( mol \right)$

   ${{C}_{6}}{{H}_{6}}+3{{H}_{2}}\underrightarrow{Ni,{{t}^{0}}} & {{C}_{6}}{{H}_{12}}$

     0,1 $\to $0,3                             (mol)

Theo phương trình ta có:

${{n}_{{{H}_{2}}}}=3.{{n}_{{{C}_{6}}{{H}_{6}}}}=3.0,1=0,3\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,3=6,72\left( l \right)$

$\Rightarrow $Chọn đáp án B