To solve this problem, we need to determine the ratio of kinetic energy to potential energy of a body when it has fallen through a distance of \frac{2h}{3} after being dropped from a height h.

1. **Initial Conditions**:

• The body is initially at rest at height h with total potential energy and no kinetic energy.
• Initial Potential Energy (PE_initial) = mgh, where m is the mass and g is the acceleration due to gravity.

2. **When the body has fallen through \frac{2h}{3}**:

• The new height is \frac{h}{3} (since the body has descended \frac{2h}{3}).
• Potential Energy at this point (PE_new) = mg \times \frac{h}{3} = \frac{mgh}{3}.

3. **Calculating the Kinetic Energy**:

• By the conservation of energy, the total mechanical energy at any point is equal to the initial potential energy.
• Total mechanical energy = Initial Potential Energy = mgh.
• Kinetic Energy (KE) when the body has fallen = Total mechanical energy - New Potential Energy.
• Kinetic Energy = mgh - \frac{mgh}{3} = \frac{2mgh}{3}.

4. **Determining the Ratio**:

• The ratio of Kinetic Energy to Potential Energy at \frac{2h}{3} height is:
• \frac{KE}{PE} = \frac{\frac{2mgh}{3}}{\frac{mgh}{3}} = \frac{2}{1}.

Thus, the ratio of the kinetic energy to potential energy when the body has fallen through a distance of \frac{2h}{3} is 2:1.