Let the hanging mass be $M$ and the block on the horizontal ice slab be $4M$.

A force of magnitude $2Mg$ pulls the bigger block horizontally.

We need the tension $T$ in the string, given that $T=\frac{x}{5}Mg.$

1. Choose acceleration direction

Since the block $4M$ is pulled strongly on the horizontal surface, the system will accelerate such that:

• block $4M$ moves horizontally,
• hanging mass $M$ moves vertically.

Let the common magnitude of acceleration be $a$.

2. Apply Newton's second law to the block $4M$

For the $4M$ block on the frictionless horizontal surface:

• external pulling force to the right = $2Mg$
• tension $T$ opposes it

So, $2Mg - T = 4Ma \quad ...(1)$

3. Apply Newton's second law to the hanging mass $M$

As the $4M$ block moves to the right, the hanging mass moves upward.

For mass $M$:

• tension upward = $T$
• weight downward = $Mg$

Taking upward as positive, $T - Mg = Ma \quad ...(2)$

4. Solve the equations

From (2), $T = Mg + Ma.$

Substitute into (1): $2Mg - (Mg + Ma) = 4Ma$ $Mg - Ma = 4Ma$ $Mg = 5Ma$ $a = \frac{g}{5}.$

Now put this in (2): $T = Mg + M\left(\frac{g}{5}\right)$ $T = \frac{6}{5}Mg.$

Comparing with $T=\frac{x}{5}Mg,$ we get $x=6.$

5. Final answer

$\boxed{6}$

The derived answer matches the stored correct answer.