Chọn D

Từ M  kẻ $MN\text{//}AB.$   Lúc đó, mặt phẳng    $\left(\alpha \right)\equiv \left(ABM\right)\equiv \left(ABMN\right).$

Ta có:   $\frac{V_{S.AMB}}{V_{S.ACB}}=\frac{SM}{SC}=\frac{1}{2}\iff V_{S.AMB}=\frac{1}{2}{V}_{S.ACB}.$

   $\frac{V_{S.AMN}}{V_{S.ACD}}=\frac{SN}{SD}.\frac{SM}{SC}=\frac{1}{4}\iff V_{S.AMN}=\frac{1}{4}{V}_{S.ACD}.$      

Suy ra:  $V_{S.AMB}+V_{S.AMN}=\frac{1}{2}{V}_{S.ABC}+\frac{1}{4}{V}_{S.ACD}.$

Mà   $V_{S.ACB}=V_{S.ACD}=\frac{1}{2}{V}_{S.ABCD}.$

Suy ra:   $V_{S.MNAB}=\frac{1}{4}{V}_{S.ABCD}+\frac{1}{8}{V}_{S.ABCD}=\frac{3}{8}{V}_{S.ABCD}=V_1.$

 $\Rightarrow V_{MNABCD}=\frac{5}{8}{V}_{S.ABCD}=V_2.$

Vậy  $\frac{V_1}{V_2}=\frac{3}{5}.$