2022-I-13

$\begin{array}{rcl}
\dfrac{\text{surface area of the smaller sphere}}{\text{surface area of the larger sphere}} & = & \left( \dfrac{\text{radius of the smaller sphere}}{\text{radius of the larger sphere}}\right)^2 \\
\dfrac{4}{9} & = &\left( \dfrac{\text{radius of the smaller sphere}}{9} \right)^2 \\
\text{radius of the smaller sphere} & = & 9 \times \dfrac{2}{3} \\
\text{radius of the smaller sphere} & = & 6 \text{ cm}
\end{array}$

Therefore, the volume of the smaller sphere

$\begin{array}{cl}
= & \dfrac{4}{3} \pi (6)^3 \\
= & 288\pi \text{ cm}^3
\end{array}$

$\begin{array}{cl}
= & 288\pi +\dfrac{4}{3} \pi (9)^3 \\
= & 1260\pi \text{ cm}^3
\end{array}$

The volume of $A$

$\begin{array}{cl}
= & \dfrac{1}{3} \pi (6)^2 (10) \\
= & 120\pi \text{ cm}^3
\end{array}$

Hence, we have

$\begin{array}{rcl}
\dfrac{\text{volume of $A$}}{\text{volume of $B$}} & = & \dfrac{120\pi}{1260\pi -120\pi} & = & \dfrac{2}{19}
\end{array}$

On the other hand,

$\begin{array}{rcl}
\left(\dfrac{\text{base radius of $A$}}{\text{base radius of $B$}}\right)^3 & = & \left(\dfrac{6}{12}\right)^3 & = & \dfrac{1}{8}
\end{array}$

Since $\dfrac{\text{volume of $A$}}{\text{volume of $B$}} \neq \left(\dfrac{\text{base radius of $A$}}{\text{base radius of $B$}}\right)^3$, then $A$ and $B$ are not similar.

The claim is not correct.