3. Let $a>b>0$. Then, the minimum value of $a^{2}+\frac{1}{b(a-b)}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
3.C.

Given $a>b>0$, we know that $0<b(a-b) \leqslant \frac{1}{4} a^{2}$. Therefore, $a^{2}+\frac{1}{b(a-b)} \geqslant a^{2}+\frac{4}{a^{2}} \geqslant 4$.