3. The base area of the rectangular prism is 4, and the length of the diagonal is
4. Then the maximum value of the lateral surface area of the rectangular prism is $\qquad$
$3.16 \sqrt{2}$.
As shown in Figure 10, let the length of the rectangular prism be $x$, the width be $y$, and the height be $z$. According to the problem, we have
$$
\left\{\begin{array}{l}
x y=4 \\
x^{2}+y^{2}+z^{2}=16 .
\end{array}\right.
$$

Thus, $z^{2}=16-\left(x^{2}+y^{2}\right) \leqslant 16-2 x y=8$.
This gives $0<z \leqslant 2 \sqrt{2}$.
And when $x=y=2$, $z=2 \sqrt{2}$. At this point,
$$
\begin{aligned}
S_{\text {worm }} & =2(x+y) z=2 z \sqrt{x^{2}+y^{2}+2 x y} \\
& =2 z \sqrt{\left(16-z^{2}\right)+8}=2 z \sqrt{24-z^{2}} \\
& =2 \sqrt{z^{2}\left(24-z^{2}\right)}=2 \sqrt{144-\left(z^{2}-12\right)^{2}} .
\end{aligned}
$$

At this time, $S_{m}$ reaches its maximum value, $S_{m}=2(2+2) \times 2 \sqrt{2}=16 \sqrt{2}$.