Given figure is cuboid then the acute angle between the diagonal OA a…

$\overrightarrow{O A}=a \hat{i}+a \hat{j}+a \hat{k} ~~~~|\overrightarrow{OA}|=\sqrt{a^2+a^2+a^2}=\sqrt{3} a$

$\overrightarrow{B E}=a \hat{i}-a \hat{j}-a \hat{A} ~~~~|\overrightarrow{B E}|=\sqrt{3}a$

So $|\overrightarrow{O A} . \overrightarrow{B E}|=|\overrightarrow{O A}||\overrightarrow{B E}| \cos \theta$

$\Rightarrow \cos \theta=\left|\frac{a^2-a^2-a^2}{a \sqrt{3} \times a \sqrt{3}}\right| =\frac{1}{3}$

so  $\theta=\cos ^{-1} \frac{1}{3}$

Option: D