3. Given in $\triangle A B C$, $A B=A C=4 \sqrt{3}$, and the altitude $A D=4$. Then the circumradius of $\triangle A B C$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
3. D.

As shown in Figure 3, extend $A D$ to intersect the circumcircle $\odot O$ of $\triangle A B C$ at point $E$. Since $A D$ is also a median, $A E$ is the diameter of $\odot O$. Connecting $B E$, we find that $\triangle A B E$ is a right triangle. Therefore, $A B^{2}=A D \cdot A E$,

which means $(4 \sqrt{3})^{2}=4 A E \Rightarrow A E=12$.