2. Given that the surface area of a rectangular prism is $\frac{45}{2}$, and the sum of the lengths of all its edges is 24, then the maximum value of the angle formed between the body diagonal of the rectangular prism and its edges is
A. $\arccos \frac{1}{3}$
B. $\arccos \frac{\sqrt{2}}{3}$
C. $\arccos \frac{\sqrt{3}}{9}$
D. $\arccos \frac{\sqrt{6}}{9}$
Let the three edge lengths of the rectangular prism be $a, b, c$, and the length of the diagonal be $l$, then $\left\{\begin{array}{l}2(a b+b c+c a)=\frac{45}{2}, \\ 4(a+b+c)=24\end{array}\right.$
$$
\Rightarrow\left\{\begin{array}{l}
b+c=6-a, \\
b^{2}+c^{2}=\frac{27}{2}-a^{2}
\end{array} \Rightarrow(6-a)^{2} \leqslant 27-2 a^{2} \Rightarrow a^{2}-4 a+3 \leqslant 0 \Rightarrow 1 \leqslant a \leqslant 3 .\right.
$$

By the symmetry of $a, b, c$, we have $1 \leqslant a, b, c \leqslant 3$, thus the cosine of the angle $\theta$ formed between the body diagonal of the rectangular prism and its edge is $\cos \theta \geqslant \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}}}=\sqrt{\frac{2}{27}}=\frac{\sqrt{6}}{9} \Rightarrow \theta \leqslant \arccos \frac{\sqrt{6}}{9}$, so the answer is $D$.