jmc

While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$ Since $GH=14,$ then $JG=14-8=6.$ So, the area of $△BJG$ is $\frac{8\cdot 6}{2}=24.$ If we let $△BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot 8=\boxed{192}.$