for a cone
$\mathrm{r}$ will be cone slant height
$$
V=2 \sqrt{3}
$$
Let $x$ radius of cone
$\mathrm{h}$ be height then
$$
\begin{array}{l}
\frac{1}{3} \pi r^{2} h=2 \sqrt{3} \pi \\
x^{2} h=6 \sqrt{3} \Rightarrow x^{2}=\frac{6 \sqrt{3}}{h}
\end{array}
$$
least Diameter $\Rightarrow$ least slant height of cone $\ell^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$ $\mathrm{r}^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$
$$
r^{2}=\frac{6 \sqrt{3}}{h}+h^{2}
$$
Diff. w. to $\mathrm{x}$
$$
2 r \frac{\mathrm{dr}}{\mathrm{dh}}=\frac{-6 \sqrt{3}}{\mathrm{~h}^{2}}+2 \mathrm{~h}
$$
for maximum \& minimum $\frac{\mathrm{dr}}{\mathrm{dh}}=0$
$$
\begin{array}{l}
\mathrm{h}=\sqrt{3} \Rightarrow \mathrm{h}^{2}=3 \\
\mathrm{x}^{2}=\frac{6 \sqrt{3}}{\sqrt{3}} \\
\mathrm{x}^{2}=6 \\
\mathrm{r}^{2}=6+3 \\
\mathrm{r}^{2}=9 \\
\mathrm{r}=3 \\
\mathrm{~d}=2 \mathrm{r} \\
\mathrm{d}=6
\end{array}
$$