3. Given that $b$ is a positive number, $a$ is the fractional part of $b$, and $a^{2}+b^{2}=27$. Then the value of $a b$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3. A.

Let $b=a+m\left(m \in \mathbf{N}_{+}, 0m^{2} \Rightarrow m4 .
\end{array}
$

Therefore, $4<m<3 \sqrt{3} \Rightarrow m=5$.
Thus, $5^{2}+10 a+2 a^{2}=27$.
Solving for $a$, we get $a=\frac{-5+\sqrt{29}}{2}$ (negative value discarded).
Hence, $a+b=a+a+5=2 a+5=\sqrt{29}$.
Therefore, $a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right]=1$.