3. As shown in Figure 2, in two concentric circles with center $O$, $M N$ is the diameter of the larger circle, intersecting the smaller circle at points $P$ and $Q$. The chord $M C$ of the larger circle intersects the smaller circle at points $A$ and $B$. If $O M=2$, $O P=1$, and $M A=A B=B C$, then the area of $\triangle M B Q$ is
3. $\frac{3 \sqrt{15}}{8}$.

Let $M A=x$.
From $M A \cdot M B=M P \cdot M Q$, we get $x \cdot 2 x=1 \times 3$.
Solving for $x$ yields $x=\sqrt{\frac{3}{2}}$.
Connect $C N$. In the right triangle $\triangle M C N$,
$$
M C=3 x=3 \sqrt{\frac{3}{2}}, M N=4 \text {. }
$$

Therefore, $N C=\sqrt{N M^{2}-M C^{2}}=\sqrt{\frac{5}{2}}$,
$$
S_{\triangle M C N}=\frac{1}{2} M C \cdot C N=\frac{3 \sqrt{15}}{4} \text {. }
$$

Also, $\frac{S_{\triangle M O B}}{S}=\frac{1}{2}$, so $S_{\triangle M Q B}=\frac{3 \sqrt{15}}{8}$.