Let $r_1 \text{ cm}$ and $r_2 \text{ cm}$ be the radius of the two hemispheres respectively with $r_1 < r_2$.
Let $A_1 \text{ cm}^2$ and $A_2 \text{ cm}^2$ be the total surface area of the two solid hemispheres respectively with $A_1 < A_2$.
Note that the hemispheres are similar to each other. Then, we have
$\begin{array}{rcl}
\dfrac{A_1}{A_2} & = & \left(\dfrac{r_1}{r_2}\right)^2 \\
\dfrac{A_1}{A_2} & = & \left(\dfrac{2}{3}\right)^2 \\
A_1 : A_2 & = & 4 : 9
\end{array}$
Therefore, we have
$\begin{array}{rcl}
A_1 & = & 351\pi \times \dfrac{4}{ 4+9} \\
A_1 & = & 108\pi \text{ cm}^2
\end{array}$
Hence, we have
$\begin{array}{rcl}
2\pi r_1^2 +\pi r_1^2 & = & 108 \pi \\
r_1^2 & = & 36 \\
r_1 & = & 6 \text{ cm}
\end{array}$
Therefore, $r_2=9\text{ cm}$.
Hence, the difference of the volumes of the two hemispheres
$\begin{array}{cl}
= & \dfrac{1}{2} \times \dfrac{4}{3} \pi (9)^3 -\dfrac{1}{2} \times \dfrac{4}{3} \pi (6)^3 \\
= & 342\pi \text{ cm}^3
\end{array}$
