Step 1:
We have
$\sin 2A+\sin 2B-\sin 2C$
$\Rightarrow 2\sin \big(\large\frac{2A+2B}{2}\big)$$\cos\big(\large\frac{2A-2B}{2}\big)$$-\sin 2C$
$\Rightarrow 2\sin(A+B)\cos(A-B)-\sin 2C$
$\sin 2C=2\sin C\cos C$
$\Rightarrow 2\sin C\cos(A-B)-2\sin C\cos C$
$\Rightarrow 2\sin C[\cos(A-B)-\cos C]$
Step 2:
The above equation can be written as
$\Rightarrow 2\sin C[\cos(A-B)+\cos(A+B)]$
$\cos(A-B)=\cos A\cos B-\sin A\sin B$
$\cos(A+B)=\cos A\cos B+\sin A\sin B$
$\Rightarrow 2\sin C[\cos A\cos B-\sin A\sin B+\cos A\cos B+\sin A\sin B$
$\Rightarrow 2\sin C(2\cos A\cos B)$
$\Rightarrow 4\cos A\cos B\sin C$
Hence (b) is the correct answer.