3. The maximum value of the area of an inscribed triangle in a circle with radius $R$ is $\qquad$

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3. $\frac{3 \sqrt{3}}{4} R^{2}$.

Let the inscribed triangle of $\odot O$ be $\triangle A B C$.
Obviously, when $\triangle A B C$ is an acute or right triangle, the area can reach its maximum value (because if $\triangle A B C$ is an obtuse triangle, the side opposite to the obtuse angle (let's assume it is $A$) can be symmetrically transformed about the center of the circle to become $B^{\prime} C^{\prime}$).
Therefore, $S_{\triangle A B^{\prime} C^{\prime}}>S_{\triangle A B C}$.
Now, let $\angle A O B=2 \alpha, \angle B O C=2 \beta$,
$\angle C O A=2 \gamma, \alpha+\beta+\gamma=\pi$.
Then $S_{\triangle A B C}=\frac{1}{2} R^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma)$.
From the discussion, we can assume $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $y=\sin x$ is a concave function on $(0, \pi)$. Then, by Jensen's inequality, we have
$$
\frac{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma}{3} \leqslant \sin \frac{2(\alpha+\beta+\gamma)}{3}=\frac{\sqrt{3}}{2} \text {. }
$$

Therefore, $S_{\triangle A B C} \leqslant \frac{1}{2} R^{2} \times 3 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} R^{2}$.
The equality holds if and only if $\triangle A B C$ is an equilateral triangle.