Three. (20 points) Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube with edge length 2, and point $M$ is the midpoint of edge $A A_{1}$. A sphere is constructed passing through points $M, B_{1}, C, D_{1}$. Try to find the radius $R$ of this sphere.
Three, as shown in Figure 3, connect $B_{1} C$, $C D_{1}$, and $D_{1} B_{1}$. Let the body diagonal $A C_{1}$ intersect the plane of the equilateral $\triangle B_{1} C D_{1}$ at point $O_{1}$. Clearly, point $O_{1}$ is the center (centroid) of the equilateral $\triangle B_{1} C D_{1}$ (by symmetry), and $A C_{1} \perp$ the plane $B_{1} C D_{1}$.

Connect $A_{1} B$, $B D$, and $D A_{1}$. Let $A C_{1}$ intersect the plane of the equilateral $\triangle A_{1} B D$ at point $O_{2}$. Since $D B \parallel D_{1} B_{1}$, $A_{1} B \parallel C D_{1}$, and $D A_{1} \parallel B_{1} C$, the plane $A_{1} B D \parallel$ the plane $B_{1} C D_{1}$.

Let $O$ be the center of the square $A B C D$. The plane $A_{1} B D$ bisects the line segment $A C$ at point $O$. Connect $O O_{2}$, then $O O_{2} \parallel C O_{1}$. Thus, $A O_{2} = O_{2} O_{1}$. Similarly, $C_{1} O_{1} = O_{1} O_{2}$.
Hence, $C_{1} O_{1} = \frac{1}{3} A C_{1} = \frac{2 \sqrt{3}}{3}$.
Let the center of the sphere be $K$, then point $K$ lies on the body diagonal $A C_{1}$.
Let $A K = x$, $\angle A_{1} A C_{1} = \theta$. By the cosine rule, we have
$$
\begin{array}{l}
R^{2} = M K^{2} = 1^{2} + x^{2} - 2 \times 1 \times x \cos \theta \\
= 1 + x^{2} - 2 x \times \frac{\sqrt{3}}{3} = x^{2} - \frac{2 \sqrt{3}}{3} x + 1 .
\end{array}
$$

It is easy to see that $A O_{1} = \frac{2}{3} A C_{1} = \frac{2}{3} \times 2 \sqrt{3} = \frac{4 \sqrt{3}}{3}$,
$$
K O_{1} = \left| \frac{4 \sqrt{3}}{3} - x \right|, O_{1} B = \frac{B_{1} C}{2 \sin 60^{\circ}} = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{6}}{3} \text{. }
$$

By the Pythagorean theorem, we have
$$
R^{2} = K B_{1}^{2} = K O_{1}^{2} + O_{1} B^{2} = \left( \frac{4 \sqrt{3}}{3} - x \right)^{2} + \frac{8}{3} \text{. }
$$

Thus, $x^{2} - \frac{2 \sqrt{3}}{3} x + 1 = \frac{16}{3} + x^{2} - \frac{8 \sqrt{3}}{3} x + \frac{8}{3}$, which simplifies to
$$
x = \frac{7 \sqrt{3}}{6} \text{. }
$$

Therefore, $R = \sqrt{\left( \frac{4 \sqrt{3}}{3} - \frac{7 \sqrt{3}}{6} \right)^{2} + \frac{8}{3}}$
$$
= \sqrt{\left( \frac{\sqrt{3}}{6} \right)^{2} + \frac{8}{3}} = \frac{\sqrt{11}}{2} \text{. }
$$