3. The set
$$
A=\{\sqrt[n]{n} \mid n \in \mathbf{N} \text { and } 1 \leqslant n \leqslant 2020\}
$$

has the maximum element as $\qquad$ .
3. $\sqrt[3]{3}$.

From $f(x)=\frac{\ln x}{x}$ being monotonically increasing on the interval $(0, \mathrm{e})$ and monotonically decreasing on the interval $(\mathrm{e},+\infty)$, we know that
$$
\begin{array}{l}
\frac{\ln 3}{3}>\frac{\ln 4}{4}>\cdots>\frac{\ln 2020}{2020} \\
\Rightarrow \ln \sqrt[3]{3}>\ln \sqrt[4]{4}>\cdots>\ln \sqrt[2020]{2020} \\
\Rightarrow \sqrt[3]{3}>\sqrt[4]{4}>\cdots>\sqrt[2020]{2020} .
\end{array}
$$

Also, $\sqrt[3]{3}>\sqrt{2}>1$, therefore, the maximum element of set $A$ is $\sqrt[3]{3}$.