$$\frac{R(\sin 2A+\sin 2B+\sin 2C)}{a+b+c}$$ $$=\frac{R(4\sin A \sin B\sin C)}{2R(\sin A+\sin B+\sin C}$$ $$=\frac{8R\sin \frac A2 \sin \frac B2 \sin \frac C2 \cos \frac A2 \cos \frac B2 \cos \frac C2}{8R\cos \frac A2 \cos \frac B2 \cos \frac C2}$$ $$=\frac{2r}{8R}$$ $$=\frac {r}{4R}$$ But the answer given is $\frac{r}{R}$
R means the circumradius of the triangle
r is the inradius
In the second line, you have a factor of $4$ in the numerator, so the numerator in the third line should be $4 \times (2 \times 2 \times 2) = 32$. The rest is fine.