Solution:

Let $A B C$ be the metallic cone, $DECB$ is the required frustum
Let the two radii of the frustum be
$\mathrm{DO}^{\prime}=\mathrm{r}_{2}$ and $\mathrm{BO}=\mathrm{r}_{1}$
From $\triangle \mathrm{ADO}^{\prime}$ and $\triangle \mathrm{ABO}$
$\begin{array}{l}
r_{2}=h_{1}\left(\tan 30^{\circ}\right)=10 \times \frac{1}{\sqrt{3}}=\frac{10}{\sqrt{3}} \\
r_{1}=\left(h_{1}+h_{2}\right) \tan 30^{\circ} \\
\quad=20 \tan 30^{\circ}=\frac{20}{\sqrt{3}}
\end{array}$
Volume of the frustum DECB $=\frac{\pi h_{2}}{3}\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)$
$\begin{array}{l}
=\frac{\pi(10)}{3} \times\left(\frac{400}{3}+\frac{200}{3}+\frac{100}{3}\right) \\
=\frac{\pi(10)}{3} \times \frac{700}{3}=\frac{7000 \pi}{9}
\end{array}$
Let $/$ be the length of the wire,
Diameter of the wire $d=\frac{1}{16} \mathrm{~cm}$,
$\therefore$ Radius of the wire $(\mathrm{R})=\frac{1}{32} \mathrm{~cm}$
$\therefore$ Volume of the frustum = Volume of the wire drawn from it
$\begin{array}{l}
\frac{7000 \pi}{9}=\mathrm{R}^{2} I \\
l=\frac{7000 \times 32 \times 32}{9} \\
\quad=796444.44 \mathrm{~cm} \text { or } 7964.44 \mathrm{~m}
\end{array}$
$\therefore$ The length of the wire $=7964.44 \mathrm{~m}$.