${{n}_{{{H}_{2}}O}}=\frac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\frac{1,62}{18}=0,09mol$

$C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O$

0,045$\leftarrow $ 0,09       $\leftarrow $ 0,09

${{m}_{{{O}_{2}}}}=0,09.32=2,88$

$a={{m}_{{{O}_{2}}}}+{{m}_{C{{H}_{4}}}}=0,09.32+0,045.16=3,6$

$%{{O}_{2}}=\frac{{{m}_{{{O}_{2}}}}}{{{m}_{hh}}}.100=\frac{2,88}{3,6}.100=80%$

                     $\Rightarrow $ Chọn đáp án A.