20. The differentiable function $f:\left(0, \frac{\pi}{2}\right) \rightarrow \mathbf{R}$ satisfies $f^{\prime}(x) \tan x \geqslant(2 \sin x-1) f(x)$, if $f\left(\frac{\pi}{6}\right)=1$, then the minimum value of $f\left(\frac{\pi}{3}\right)$ is . $\qquad$
$$
\begin{array}{l}
f^{\prime}(x) \tan x \geqslant(2 \sin x-1) f(x) \Rightarrow f^{\prime}(x) \sin x \geqslant(2 \sin x \cos x-\cos x) f(x) \\
\Rightarrow f^{\prime}(x) \sin x+f(x) \cos x \geqslant 2 \cos x \cdot f(x) \sin x .
\end{array}
$$

Let $g(x)=f(x) \sin x, x \in\left(0, \frac{\pi}{2}\right)$, and let $G(x)=\mathrm{e}^{-2 \sin x} \cdot g(x)$, then $g^{\prime}(x)-2 \cos x \cdot g(x) \geqslant 0$ $\Rightarrow \mathrm{e}^{-2 \sin x}\left[g^{\prime}(x)-2 \cos x \cdot g(x)\right] \geqslant 0 \Rightarrow G^{\prime}(x) \geqslant 0 \Rightarrow G(x)$ is monotonically increasing, thus
$$
G\left(\frac{\pi}{3}\right) \geqslant G\left(\frac{\pi}{6}\right) \Rightarrow \mathrm{e}^{-\sqrt{3}} f\left(\frac{\pi}{3}\right) \cdot \frac{\sqrt{3}}{2} \geqslant \mathrm{e}^{-1} f\left(\frac{\pi}{6}\right) \cdot \frac{1}{2} \Rightarrow \sqrt{3} \mathrm{e}^{-\sqrt{3}} f\left(\frac{\pi}{3}\right) \geqslant \mathrm{e}^{-1} \Rightarrow f\left(\frac{\pi}{3}\right) \geqslant \frac{\mathrm{e}^{\sqrt{3}-1}}{\sqrt{3}},
$$

When the equality holds, $G(x)=G\left(\frac{\pi}{6}\right)=\frac{1}{2} \mathrm{e}^{-1} \Rightarrow \mathrm{e}^{-2 \sin x} \cdot f(x) \sin x=\frac{1}{2} \mathrm{e}^{-1} \Rightarrow f(x)=\frac{\mathrm{e}^{2 \sin x-1}}{2 \sin x}$. Therefore, the minimum value of $f\left(\frac{\pi}{3}\right)$ is $\frac{\mathrm{e}^{\sqrt{3}-1}}{\sqrt{3}}$.