2. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=4, A A_{1}=A D=2$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, B C$ respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$
2.3.

Take point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{2}$. Then $H E / / B_{1} G$. Therefore, $H E / /$ plane $B_{1} F G$.
Thus, $V_{B_{1}-E F G}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H}$.
And $S_{\triangle B_{1} F H}=\frac{9}{2}$, the distance from point $G$ to plane $B_{1} F H$ is 2. Hence $V_{B_{1}-E F G}=3$.