6. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and $M, N$ are the midpoints of edges $C C_{1}$ and $A_{1} D_{1}$, respectively. Then the volume of the tetrahedron $O-$ $M N B_{1}$ is ( ).
(A) $\frac{1}{6}$
(B) $\frac{5}{48}$
(C) $\frac{1}{8}$
(D) $\frac{7}{48}$
6. (D):

As shown in Figure 3, draw a line through $O$ parallel to $AB$, intersecting edges $AD$ and $BC$ at points $E$ and $F$ respectively. Connect $BE$, and take the midpoint of $BF$ as $Q$. It is easy to see that $OQ \parallel BE \parallel B_{1}N$. Therefore, $OQ \parallel$ plane $MNB_{1}$. Hence, $V_{O M N B_{1}} \doteq$
$$
\begin{array}{l}
V_{Q-M N B_{1}}=V_{N-M Q B_{1}}. \text{ Also, } S_{\triangle M Q B_{1}}=1-\frac{1}{4}-\frac{3}{16}-\frac{1}{8} \\
=\frac{7}{16} \text{, so, } V_{O M B_{1}}=\frac{7}{48} .
\end{array}
$$