The diameter of a metallic sphere is equal to 9 cm.

The radius of the metallic sphere is $\frac{9}{2}=4.5 \mathrm{~cm}=45 \mathrm{~mm}$. Therefore, the volume of the metallic sphere is

$V=\frac{4}{3} \pi \times(45)^{3}$ Cubic $\mathrm{mm}$

The metallic sphere is melted to produce a long wire of uniform cross section of radius $\frac{2}{2}=1 \mathrm{~mm}$. Let the length of the wire be / $\mathrm{mm}$. Then, the volume of the wire is

$V_{1}=\pi \times(1)^{2} \times l=\pi l$ Cubic $\mathrm{mm}$

Since, the volume of the metallic sphere is equal to the volume of the wire, we have

$V=V_{1}$

$\Rightarrow \frac{4}{3} \pi \times(45)^{3}=\pi l$

$\Rightarrow \quad l=\frac{4}{3} \times(45)^{3}$

$\Rightarrow \quad=4 \times(45)^{2} \times 15$

$\Rightarrow \quad=121500$

Hence, the length of the wire is $121500 \mathrm{~mm}=12150 \mathrm{~cm}$.

Hence length $=12150 \mathrm{~cm}$