3. If $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a unit cube, and $M$ is the midpoint of edge $B B^{\prime}$, then the distance from point $M$ to the plane $A^{\prime} C^{\prime} D$ is ( ).
(A) 1
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\sqrt{3}-1$
3. C.

As shown in Figure 2, connect $B D$, $B A^{\prime}$, $B C^{\prime}$, and $B^{\prime} D$.

Since $M$ is the midpoint of $B B^{\prime}$, the distance from point $M$ to the plane $A^{\prime} C^{\prime} D$ is half the sum of the distances from point $B$ to the plane $A^{\prime} C^{\prime} D$ and from point $B^{\prime}$ to the plane $A^{\prime} C D$.
Since $V_{B-1 C^{\prime} y}=1^{3}-4 \times \frac{1}{3} \times \frac{1}{2} \times 1^{3}=\frac{1}{3}$,
$V_{B^{\prime}-A C^{\prime} D}=\frac{1}{3} S_{\triangle M F C^{\prime}} \cdot D D^{\prime}=\frac{1}{3} \times \frac{1}{2} \times 1^{3}=\frac{1}{6}$,
thus, $V_{n-x C^{c} p}=\frac{V_{n-s c^{\prime} D}+V_{U r-1} C^{c} p}{2}=\frac{1}{4}$.
And $S_{\triangle S^{\prime} C^{\prime} D}=\frac{1}{2}\left(A^{\prime} C^{\prime}\right)^{2} \sin 60^{\circ}=\frac{\sqrt{3}}{2}$, so the distance from point $M$ to the plane $A^{\prime} C^{\prime} D$ is
$$
\frac{V_{M-A C^{\prime} D}}{\frac{1}{3} S_{\triangle C^{\prime} C^{\prime} D}}=\frac{\sqrt{3}}{2} .
$$