We will solve this problem step by step using the principles of energy conservation and work-energy theorem.

### Given Data:
- Mass of the object, \( m = 234 \) g \( = 0.234 \) kg
- Height from which the object is released, \( h = 1.05 \) m
- Length of the flat part, \( L = 2.16 \) m
- Energy lost due to friction per traversal of the flat section, \( E_{\text{loss}} = 688 \) mJ \( = 0.688 \) J
- Acceleration due to gravity, \( g = 9.81 \) m/s²

### Step 1: Find the Initial Mechanical Energy
The total initial energy of the object at point A consists of its gravitational potential energy (since it starts from rest, kinetic energy is zero):

\[
E_{\text{initial}} = mgh
\]

Substituting the values:

\[
E_{\text{initial}} = (0.234)(9.81)(1.05)
\]

\[
E_{\text{initial}} = 2.408 J
\]

### Step 2: Energy Consideration on the Track
- The object starts at \( A \) with 2.408 J of energy.
- It reaches the flat part and moves with some speed.
- Each time it traverses the flat part, it loses 0.688 J due to friction.
- The object will keep moving back and forth until all mechanical energy is dissipated by friction.

### Step 3: Finding the Number of Complete Traversals
Each time the object crosses the flat part completely, it loses 0.688 J. We need to find how many complete crossings it can make before all the energy is dissipated.

Let \( n \) be the number of times the object completely traverses the flat part:

\[
n \times 0.688 = 2.408
\]

\[
n = \frac{2.408}{0.688} = 3.5
\]

This means the object completes **3 full traversals** and then moves halfway across before stopping.

### Step 4: Final Stopping Position
- The object moves back and forth across the flat region.
- After 3 full trips, it still has \( 0.408 \) J of energy left.
- Since one full crossing costs 0.688 J, a remaining 0.408 J means it travels partially across the flat section before stopping.

To find how far it moves in the final traversal, we use the work-energy principle:

\[
\text{Work done by friction} = \text{Remaining Energy} = 0.408 J
\]

Frictional force \( f \) is given by:

\[
f = \frac{E_{\text{loss}}}{L} = \frac{0.688}{2.16} = 0.319 J/m
\]

The stopping distance \( d \) is:

\[
d = \frac{0.408}{0.319} = 1.28 \text{ m}
\]

### Final Answer:
After 3 complete traversals, the object moves **1.28 m on its last trip before stopping**. It comes to rest **1.28 m from the left end of the flat section**.