3. Given that the surface area of a rectangular prism is $\frac{45}{2} \mathrm{~cm}^{2}$, and the total length of all edges is $24 \mathrm{~cm}$. Then, the maximum angle formed between the body diagonal of the rectangular prism and an edge is $($.
(A) $\arccos \frac{1}{3}$
(B) $\arccos \frac{\sqrt{2}}{3}$
(C) $\arccos \frac{\sqrt{3}}{9}$
(D) $\arccos \frac{\sqrt{6}}{9}$
3. D.

Let the lengths of the three edges meeting at the same vertex be $a, b, c (a \geqslant b \geqslant c)$. Then
$$
\left\{\begin{array}{l}
2(a b + b c + c a) = \frac{45}{2} \\
4(a + b + c) = 24
\end{array} \Rightarrow \left\{\begin{array}{l}
a + b = 6 - c, \\
a b = c^{2} - 6 c + \frac{45}{4}.
\end{array}\right.\right.
$$

Thus, $a, b$ are the roots of the equation $x^{2} - (6 - c)x + c^{2} - 6c + \frac{45}{4} = 0$.
Therefore, $\Delta = (6 - c)^{2} - 4\left(c^{2} - 6c + \frac{45}{4}\right) \geqslant 0$.
Solving this, we get $1 \leqslant c \leqslant 3$.
Since $c \leqslant \frac{1}{3}(a + b + c) = 2$, we have $1 \leqslant c \leqslant 2$.
$$
\begin{array}{l}
\text{Also, } a^{2} + b^{2} + c^{2} \\
= (a + b + c)^{2} - 2(a b + b c + c a) = \frac{27}{2} \\
\Rightarrow \cos \theta = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{\sqrt{6}}{9} c \\
\Rightarrow \frac{\sqrt{6}}{9} \leqslant \cos \theta \leqslant \frac{2 \sqrt{6}}{9} \\
\Rightarrow \arccos \frac{2 \sqrt{6}}{9} \leqslant \theta \leqslant \arccos \frac{\sqrt{6}}{9}.
\end{array}
$$