Step-by-Step Solution:

1. Analyze the Initial State:

   • Initially, the cylinder of height $L_0$ is open at the bottom and has a small hole at the top. The piston is removed. This means the air inside the cylinder is at atmospheric pressure, $P_0$.
   • The hole at the top is then sealed. At this moment, we have a column of air of height $L_0$ trapped inside the cylinder.
   • The initial pressure of the trapped air is $P_1 = P_0$.
   • The initial volume of the trapped air is $V_1 = A L_0$, where $A = \pi R^2$ is the cross-sectional area of the cylinder.
   • The information about the piston (mass $M$ and position $L$) is extraneous to the final equilibrium state.

2. Analyze the Final State:

   • The cylinder is placed in a water tank such that the water level in the tank is at the same level as the top of the cylinder.
   • Water enters the cylinder from the bottom, compressing the trapped air. In equilibrium, the water has risen to a height $H$ inside the cylinder.
   • The height of the trapped air column is now $L_0 - H$.
   • The final volume of the trapped air is $V_2 = A(L_0 - H)$.
   • Let the final pressure of the trapped air be $P_2$.

3. Apply Boyle's Law:

   • The cylinder is described as "thermally conducting," which implies that the temperature of the trapped air remains constant throughout the process (isothermal process).
   • According to Boyle's Law for an isothermal process, $P_1V_1 = P_2V_2$.
   • Substituting the values from steps 1 and 2: $P_0 (A L_0) = P_2 (A(L_0 - H))$
   • The area $A$ cancels out: $P_0 L_0 = P_2(L_0 - H)$
   • Solving for the final pressure $P_2$: $P_2 = \frac{{{P_0}{L_0}}}{{{L_0} - H}} \quad \quad (1)$

4. Apply Hydrostatic Equilibrium:

   • In the final equilibrium state, we consider the pressure at the surface of the water inside the cylinder.
   • The pressure at this level exerted by the trapped air from above is $P_2$.
   • This pressure must be balanced by the pressure at the same horizontal level outside the cylinder.
   • The water surface in the tank is at the top of the cylinder. The water surface inside the cylinder is at a height $H$ from the bottom, which means it is at a depth of $(L_0 - H)$ from the water surface in the tank.
   • The pressure at a depth $(L_0 - H)$ below the free surface of the water in the tank is the sum of the atmospheric pressure ($P_0$) at the surface and the hydrostatic pressure of the water column of height $(L_0 - H)$. $P_{outside} = P_0 + \rho g (L_0 - H)$
   • For equilibrium, the pressure inside must equal the pressure outside at the same level: $P_2 = P_0 + \rho g(L_0 - H) \quad \quad (2)$

5. Combine and Solve for H:

   • We now have two expressions for the final pressure $P_2$. Equating equations (1) and (2): $\frac{{{P_0}{L_0}}}{{{L_0} - H}} = P_0 + \rho g(L_0 - H)$
   • To eliminate the fraction, multiply both sides by $(L_0 - H)$: ${P_0}{L_0} = P_0(L_0 - H) + \rho g(L_0 - H)(L_0 - H)$ ${P_0}{L_0} = P_0(L_0 - H) + \rho g{(L_0 - H)^2}$
   • Rearrange the terms to match the format of the given options: $\rho g{(L_0 - H)^2} + P_0(L_0 - H) - {P_0}{L_0} = 0$

6. Compare with Options:

   • The derived equation is $\rho g{({L_0} - H)^2} + {P_0}({L_0} - H) - {L_0}{P_0} = 0$.

   • This matches option C.

   • A: $\rho g{({L_0} - H)^2} + {P_0}({L_0} - H) + {L_0}{P_0} = 0$ (Incorrect sign for the last term)

   • B: $\rho g{({L_0} - H)^2} - {P_0}({L_0} - H) - {L_0}{P_0} = 0$ (Incorrect sign for the second term)

   • C: $\rho g{({L_0} - H)^2} + {P_0}({L_0} - H) - {L_0}{P_0} = 0$ (Correct)

   • D: $\rho g{({L_0} - H)^2} - {P_0}({L_0} - H) + {L_0}{P_0} = 0$ (Incorrect signs for the second and third terms)