3. In a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, it is known that $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$, $M$ is a point on the edge $B B_{1}$, and $S_{\triangle D B M}: S_{\triangle O_{1} B_{1} M}=$ $2: 3$. Then the volume of the tetrahedron $O_{1} A D M$ is $\qquad$ .
3. $\frac{7}{48}$.

As shown in Figure 2, let $O$ be the center of the base $ABCD$. Then $AO \perp$ plane $DO_1M$.
From $\frac{S_{\triangle DBM}}{S_{\triangle O_1B_1M}}=\frac{2BM}{B_1M}=\frac{2}{3} \Rightarrow \frac{BM}{B_1M}=\frac{1}{3}$.
Thus, $BM=\frac{1}{4}, B_1M=\frac{3}{4}$.
Therefore, $S_{\triangle DO_1M}$
$$
\begin{array}{l}
=S_{\text {square } D_1B_1BD}-S_{\triangle OD_1O_1}-S_{\triangle O_1B_1M}-S_{\triangle DBM} \\
=\frac{7 \sqrt{2}}{16} .
\end{array}
$$

Then $V_{\text {tetrahedron }-O_1MD}=\frac{1}{3} S_{\triangle DO_1M} \cdot AO$
$$
=\frac{1}{3} \times \frac{7 \sqrt{2}}{16} \times \frac{\sqrt{2}}{2}=\frac{7}{48} \text {. }
$$