6. As shown in Figure 2, in the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $P$, $Q$, and $R$ are the midpoints of edges $A B$, $A D$, and $A A_{1}$, respectively. A right triangular prism is constructed with $\triangle P Q R$ as its base, such that the three vertices of its other base are also on the surface of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$. The volume of this right triangular prism is ( ).
(A) $\frac{3}{8}$
(B) $\frac{\sqrt{3}}{8}$
(C) $\frac{3}{16}$
(D) $\frac{\sqrt{3}}{16}$
6. C.

It is known that the three vertices $P_{1}, Q_{1}, R_{1}$ of the other base of the right triangular prism are the centers of the squares $B C C_{1} B_{1}$, $C D D_{1} C_{1}$, and $A_{1} B_{1} C_{1} D_{1}$, respectively. Therefore, the lateral edge of the right triangular prism $P Q R-P_{1} Q_{1} R_{1}$ is $P P_{1}=\frac{1}{2} A C_{1}=\frac{\sqrt{3}}{2}$.
Also, $S_{\triangle P Q R}=\frac{\sqrt{3}}{4}\left(\frac{\sqrt{2}}{2}\right)^{2}=\frac{\sqrt{3}}{8}$, hence
$$
V_{\text {right triangular prism } P Q R-P_{1} Q_{1} R_{1}}=S_{\triangle P Q R} \cdot P P_{1}=\frac{3}{16} \text {. }
$$