A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width.

We have,

Radius of the metallic sphere, $R=\frac{8}{2}=4 \mathrm{~cm}$ and

Height of the cylindrical wire, $h=12 \mathrm{~m}=1200 \mathrm{~cm}$

Let the radius of the base be $r$.

Now,

Volume of the cylindrical wire $=$ Volume of the metallic sphere

$\Rightarrow \pi r^{2} h=\frac{4}{3} \pi R^{3}$

$\Rightarrow r^{2}=\frac{4 R^{3}}{3 h}$

$\Rightarrow r^{2}=\frac{4 \times 4 \times 4 \times 4}{3 \times 1200}$

$\Rightarrow r^{2}=\frac{16}{225}$

$\Rightarrow r=\sqrt{\frac{16}{225}}$

$\Rightarrow r=\frac{4}{15} \mathrm{~cm}$

$\therefore$ The width of the wire $=2 r$

$=2 \times \frac{4}{15}$

$=\frac{8}{15} \mathrm{~cm}$

So, the width of the wire is $\frac{8}{15} \mathrm{~cm}$.