A pond has been dug at the Indian Statistical Institute as an inverted truncated pyramid with a square base (see figure below). The depth of the pond is 6m. The square at the bottom has side length 2m and the top square has side length 8m. Water is filled in at a rate of $\tfrac{19}{3}$ cubic meters per hour. At what rate is the water level rising exactly $1$ hour after the water started to fill the pond?

[img]https://cdn.artofproblemsolving.com/attachments/0/9/ff8cac4bb4596ec6c030813da7e827e9a09dfd.png[/img]
1. **Volume of the Truncated Pyramid:**
   The volume \( V \) of a truncated pyramid with a square base can be calculated using the formula:
   \[
   V = \frac{h}{3} \left( A_1 + A_2 + \sqrt{A_1 A_2} \right)
   \]
   where \( h \) is the height, \( A_1 \) is the area of the top square, and \( A_2 \) is the area of the bottom square.

2. **Given Values:**
   - Height \( h = 6 \) meters
   - Side length of the bottom square \( s_2 = 2 \) meters, so \( A_2 = 2^2 = 4 \) square meters
   - Side length of the top square \( s_1 = 8 \) meters, so \( A_1 = 8^2 = 64 \) square meters

3. **Volume Calculation:**
   \[
   V = \frac{6}{3} \left( 64 + 4 + \sqrt{64 \cdot 4} \right) = 2 \left( 64 + 4 + 16 \right) = 2 \times 84 = 168 \text{ cubic meters}
   \]

4. **Volume of Water as a Function of Height:**
   Let \( x \) be the height of the water. The side length of the water surface at height \( x \) can be found using similar triangles:
   \[
   s(x) = 2 + \frac{6 - x}{6} \cdot (8 - 2) = 2 + x
   \]
   Therefore, the area of the water surface at height \( x \) is:
   \[
   A(x) = (2 + x)^2
   \]
   The volume of water \( V(x) \) up to height \( x \) is:
   \[
   V(x) = \frac{x}{3} \left( 4 + (2 + x)^2 + \sqrt{4 \cdot (2 + x)^2} \right) = \frac{x}{3} \left( 4 + (2 + x)^2 + 2(2 + x) \right)
   \]
   Simplifying:
   \[
   V(x) = \frac{x}{3} \left( 4 + x^2 + 4x + 4 + 2x \right) = \frac{x}{3} \left( x^2 + 6x + 8 \right) = \frac{x(x^2 + 6x + 8)}{3}
   \]

5. **Volume of Water After 1 Hour:**
   The volume of water after \( t \) hours is:
   \[
   V = \frac{19t}{3}
   \]
   Setting \( t = 1 \):
   \[
   V = \frac{19 \cdot 1}{3} = \frac{19}{3} \text{ cubic meters}
   \]

6. **Height of Water After 1 Hour:**
   Equate the volume expressions:
   \[
   \frac{x(x^2 + 6x + 8)}{3} = \frac{19}{3}
   \]
   Simplifying:
   \[
   x(x^2 + 6x + 8) = 19
   \]
   Solving for \( x \):
   \[
   x^3 + 6x^2 + 8x - 19 = 0
   \]
   This cubic equation can be solved numerically or using a suitable method to find \( x \).

7. **Rate of Change of Water Level:**
   Differentiate the volume equation implicitly with respect to time \( t \):
   \[
   \frac{d}{dt} \left( \frac{x(x^2 + 6x + 8)}{3} \right) = \frac{19}{3}
   \]
   Using the product rule:
   \[
   \frac{1}{3} \left( x^2 + 6x + 8 \right) \frac{dx}{dt} + \frac{1}{3} x \left( 2x + 6 \right) \frac{dx}{dt} = \frac{19}{3}
   \]
   Simplifying:
   \[
   \left( x^2 + 6x + 8 + x(2x + 6) \right) \frac{dx}{dt} = 19
   \]
   \[
   \left( x^2 + 6x + 8 + 2x^2 + 6x \right) \frac{dx}{dt} = 19
   \]
   \[
   \left( 3x^2 + 12x + 8 \right) \frac{dx}{dt} = 19
   \]
   Solving for \( \frac{dx}{dt} \):
   \[
   \frac{dx}{dt} = \frac{19}{3x^2 + 12x + 8}
   \]

8. **Evaluating at \( t = 1 \):**
   Substitute the value of \( x \) found from the cubic equation into the derivative to find the rate of change of the water level.

The final answer is \( \boxed{ \frac{19}{6\sqrt{3}} } \).