An aquarium on a level table has rectangular faces and is $10$ inches wide and $8$ inches high. When it was tilted, the water in it covered an $8"\times 10"$ end but only three-fourths of the rectangular room. The depth of the water when the bottom was again made level, was

$\textbf{(A) }2\textstyle{\frac{1}{2}}"\qquad\textbf{(B) }3"\qquad\textbf{(C) }3\textstyle{\frac{1}{4}}"\qquad\textbf{(D) }3\textstyle{\frac{1}{2}}"\qquad \textbf{(E) }4"$
1. **Understanding the problem**: The aquarium has a rectangular base of dimensions \(10\) inches by \(8\) inches and is \(8\) inches high. When tilted, the water covers an \(8 \times 10\) end but only three-fourths of the rectangular bottom. We need to find the depth of the water when the aquarium is level again.

2. **Visualizing the tilt**: When the aquarium is tilted, the water forms a right triangle on the \(8 \times 10\) end. The height of this triangle is \(8\) inches (the height of the aquarium), and the base of the triangle is \(\frac{3}{4} \times 10 = 7.5\) inches (since it covers three-fourths of the bottom).

3. **Calculating the area of the triangle**: The area of the water triangle when tilted is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7.5 \times 8 = 30 \text{ square inches}
   \]

4. **Finding the volume of water**: Since the area of the triangle represents the cross-sectional area of the water, and the width of the aquarium is \(10\) inches, the volume of the water is:
   \[
   \text{Volume} = \text{Area} \times \text{width} = 30 \times 10 = 300 \text{ cubic inches}
   \]

5. **Calculating the depth of water when level**: When the aquarium is level, the water will form a rectangular prism with a base area of \(10 \times 8 = 80\) square inches. Let \(h\) be the depth of the water. The volume of the water is:
   \[
   \text{Volume} = \text{base area} \times \text{depth} = 80 \times h
   \]
   Setting this equal to the volume of water calculated earlier:
   \[
   80h = 300
   \]
   Solving for \(h\):
   \[
   h = \frac{300}{80} = 3.75 \text{ inches}
   \]

6. **Conclusion**: The depth of the water when the aquarium is level is \(3.75\) inches.

The final answer is \(\boxed{3.75}\)