3. In a cube with edge length 1, construct a large inscribed sphere $O_{1}$, and then construct a smaller sphere $O_{2}$ inside the cube, such that it is externally tangent to the larger sphere and also tangent to three faces of the cube. Then the surface area of sphere $O_{2}$ is $\qquad$ .
3. $(7-4 \sqrt{3}) \pi$.

As shown in Figure 1, let
the cube be $A B C D$ $-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, connect $B D^{\prime}$, intersecting the sphere $O_{1}$ at points $E$ and $F$, with the radii of spheres $O_{1}$ and $O_{2}$ being $r_{1}$ and $r_{2}$, respectively. It is easy to see that,
Figure 1
$$
\begin{array}{l} 
r_{1}=\frac{1}{2}, \\
\frac{O_{1} B}{O_{1} F}=\frac{O_{2} B}{O_{2} E} . \\
\text { Then } \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{O_{2} B}{r_{2}}, O_{2} B+r_{2}=\frac{\sqrt{3}}{2}-\frac{1}{2} \\
\Rightarrow r_{2}=\frac{2-\sqrt{3}}{2} .
\end{array}
$$

Therefore, the surface area of sphere $O_{2}$ is $(7-4 \sqrt{3}) \pi$.