smc

Let the third dimension of the aquarium be $x$, and let the height of the water when the aquarium is level be $h$. When the aquarium is tilted, the water forms a triangular prism. The triangular faces have height $8$ and, from the problem, base $\frac{3}{4}x$. Thus, they have area $\frac{1}{2}\cdot 8\cdot \frac{3}{4}x=3x$, so, because the prism has length $10$, the volume of the water is $10\cdot 3x=30x$. When the tank is level, the water forms a rectangular prism with volume $h\cdot 10\cdot x=10hx$. Because the amount of water in the tank is conserved, we can equate these two expressions for the volume of the water. Thus, $10hx=30x$, so $h=\boxed{3"}$.