Sphere of radius $r=3 \mathrm{~cm}$
Let $b, h$ be base radius and height of cone respectively.
So, volume of cone $=\frac{1}{2} \pi b^2 h$


In right angled $\triangle A B C$ by Pythagoras theorem
$$
\begin{aligned}
&(h-r)^2+b^2=r^2 \\
\Rightarrow & b^2=r^2-(h-r)^2=r^2-\left(h^2-2 h r+\mathrm{r}^2\right)=2 h r-h^2 \\
\therefore & \text { Volume }(v) \frac{1}{3} \pi h\left[2 h r-h^2\right]=\frac{1}{3}\left[2 h r^2-h^3\right] \\
& \frac{d v}{d h}=\frac{1}{3}\left[4 h r-3 h^2\right]=0 \Rightarrow h(4 r-3 h)=0 \\
& \frac{d^2 v}{d h^2}=\frac{1}{3}[4 r-6 h]
\end{aligned}
$$
At $h=\frac{4 r}{3}, \frac{d^2 v}{d h^2}=\frac{1}{3}\left[4 r-\frac{4 r}{3} \times 6\right]=\frac{1}{3}[4 r-8 r < ] 0$
$\Rightarrow$ maximum volume ocurs at $h$
$$
=\frac{4 r}{3}=\frac{4}{3} \times 3=4 \mathrm{~cm}
$$
As from (1),
$$
(h-r)^2+b^2=r^2
$$
$$
\Rightarrow b^2=2 h r-h^2=2 \cdot \frac{4 r}{3} r-\frac{16 r^2}{9}
$$
$$
\begin{aligned}
&=\frac{8 r^2}{3}-\frac{16 r^2}{9} \\
&=\frac{(24-16) r^2}{9}=\frac{8 r^2}{9} \\
&\Rightarrow b=\frac{2 \sqrt{2}}{3} r=2 \sqrt{2} \mathrm{~cm}
\end{aligned}
$$
Therefore curved surface area
$$
\begin{aligned}
&=\pi b l=\pi b \sqrt{h^2+r^2} \\
&=\pi 2 \sqrt{2} \sqrt{4^2+8}=8 \sqrt{3} \pi \mathrm{cm}^2
\end{aligned}
$$