$\begin{cases}n_{KOH}=0,06 mol \\n_{NaOH}=0,06 mol \end{cases}\Rightarrow m_X+m_{KOH}+m_{NaOH}=m_{cr}+m_{H_2O}$$\Rightarrow m_{H_2O}=3,6+0,06.56+0,06.40-8,28=1,08 gam$
$n_X=n_{H_2O}=0,06 mol\Rightarrow M_X=\frac{3,6}{0,06}=60\Rightarrow X:CH_3COOH$
$\Rightarrow Chon:B$