3. Let $a, b \in \mathbf{N}_{+}$, and satisfy
$$
56 \leqslant a+b \leqslant 59,0.9<\frac{a}{b}<0.91 \text {. }
$$

Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182
3. B.

From the problem, we get $0.9 b+b56$.
Solving, we get $29<b<32 \Rightarrow b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we solve to get $27<a<28$, there is no such positive integer $a$;

When $b=31$, from $0.9 b<a<0.91 b$, we solve to get $27<a<29 \Rightarrow a=28$.
Thus, $b^{2}-a^{2}=177$.