3. As shown in the figure, $A, B, C$ are three points on the surface of a sphere with a surface area of $48 \pi$. $A B=2 \cdot B C=4$, $\angle A B C=60^{\circ}$, O is the center of the sphere, then the angle formed by the line $O A$ and the plane $A B C$ is
A. $\arcsin \frac{\sqrt{3}}{6}$
B. $\arccos \frac{\sqrt{3}}{6}$
C. $\arcsin \frac{\sqrt{3}}{3}$
D. $\arccos \frac{\sqrt{3}}{3}$
3. $\mathrm{D}$ As shown in the figure, let the radius of the sphere be $R$, then $48 \pi=4 \pi R^{2}$, so $R=2 \sqrt{3}$. In $\triangle A B C$, by the cosine rule we have
$$
A C=\sqrt{A B^{2}+B C^{2}-2 A B \cdot B C \cos \angle A B C}=2 \sqrt{3} .
$$

Noticing that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $A B \perp A C$, making $\triangle A B C$ a right triangle.
Since $O A=O B=O C$,
the projection of point $O$ on the plane $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $H$ of the hypotenuse $B C$, so the angle between line $O A$ and the section $A B C$ is equal to $\angle O A H$.

In the right $\triangle O H A$, $A H=\frac{1}{2} B C=2, O A=2 \sqrt{3}$,
$\cos \angle O A H=\frac{A H}{O A}=\frac{\sqrt{3}}{3}$, hence $\angle O A H=\arccos \frac{\sqrt{3}}{3}$ is the required angle.