jmc

Let the base triangles have sides $a$ and $b$ with included angle $\theta ,$ and let the right prism have altitude $h$.



Then the surface area constraint is

$$ah+bh+\frac{1}{2}ab\sin \theta =24,$$and the volume is

$$V=\frac{1}{2}abh\sin \theta .$$Let $X=ah,Y=bh,Z=(ab\sin \theta )/2$ be the areas of the three faces. Then $X+Y+Z=24$, and $$XYZ=\frac{1}{2}{a}^2{b}^2{h}^2\sin \theta =\frac{2}{\sin \theta }{\left(\frac{1}{2}abh\sin \theta \right)}^2=\frac{2V^2}{\sin \theta }.$$Now the AM-GM inequality yields

$$(XYZ{)}^{1/3}\le \frac{X+Y+Z}{3}=8,$$so $XYZ\le 512$. But $$\frac{2V^2}{\sin \theta }=XYZ\le 512,$$so $$V^2\le 256\sin \theta \le 256,$$which means $V\le 16$.

Equality occurs for $a=b=4$, $h=2$, and $\theta =\pi /2$, so the maximum volume of the prism is $\boxed{16}$.