Let the balls be numbered from the free hanging end upward along the string.

When the 6th ball just leaves the table, that means:

• balls $1$ to $6$ are hanging vertically,
• balls $7$ to $10$ are still on the smooth table.

Each ball has mass $m=2\,\text{kg}$.

1. Find acceleration of the whole system

Since the table is smooth, the only external force causing motion along the string is the weight of the hanging balls.

Hanging mass

$M_h = 6m = 6\times 2 = 12\,\text{kg}$

Total mass

$M = 10m = 10\times 2 = 20\,\text{kg}$

So the acceleration of the system is $a = \frac{M_h g}{M} = \frac{12g}{20} = \frac{3g}{5}$

Taking $g=10\,\text{m/s}^2$, $a = \frac{3\times 10}{5} = 6\,\text{m/s}^2$

2. Find tension between 7th and 8th ball

Consider the subsystem consisting of balls $8,9,10$ only.

These three balls are on the table, so the only horizontal external force on this subsystem is the tension $T$ in the string between balls $7$ and $8$.

Total mass of this subsystem: $M' = 3m = 3\times 2 = 6\,\text{kg}$

Since all balls move with acceleration $a=6\,\text{m/s}^2$, we have $T = M'a = 6\times 6 = 36\,\text{N}$

3. Final answer

$\boxed{36\,\text{N}}$

4. Comparison with stored answer

Stored correct answer = $36$

Our derived answer = $36$

So the answer matches.