3. If $[a]$ denotes the integer part of the real number $a$, then $\left[\frac{1}{\sqrt{16-6 \sqrt{7}}}\right]$ equals ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3. B.
$$
\begin{array}{l}
\because 16-6 \sqrt{7}=4+7-2 \sqrt{63} \\
\quad=(\sqrt{9}-\sqrt{7})^{2}, \\
\therefore \frac{1}{\sqrt{16-6 \sqrt{7}}}=\frac{1}{\sqrt{9}-\sqrt{7}}=\frac{3+\sqrt{7}}{2} .
\end{array}
$$

And $2<\sqrt{7}<3$,
$$
\therefore \frac{5}{2}<\frac{3+\sqrt{7}}{2}<3 \text {. }
$$

Thus $\left[\frac{1}{\sqrt{16-6 \sqrt{7}}}\right]=2$.