Interpret the setup

We have:

• A hanging mass: $m_1 = 5\,\text{kg}$
• A block on a rough horizontal surface: $m_2 = 10\,\text{kg}$
• Coefficient of friction: $\mu = 0.15$
• An extra mass $m$ is placed on top of $m_2$ so that the motion just stops.

Since the question says the blocks are moving as shown, the natural interpretation is:

• $m_1$ tends to move downward,
• $m_2$ tends to move horizontally,
• friction opposes the motion of $m_2$.

Condition to stop the motion

For the system to be just prevented from moving, acceleration must be zero.

Then for the hanging mass $m_1$: $T = m_1 g$

For the block on the horizontal surface, limiting friction must balance the pull by the string: $f = T$

Maximum friction available is: $f_{\max} = \mu N$

Since the extra mass $m$ is placed on top of $m_2$, the normal reaction is: $N = (m_2 + m)g$

Therefore, $f_{\max} = \mu (m_2 + m)g$

For minimum mass required to just stop the motion: $T = f_{\max}$

Substitute values

$m_1 g = \mu (m_2 + m)g$

Cancel $g$: $m_1 = \mu (m_2 + m)$

Put values: $5 = 0.15(10 + m)$

Solve for $m$

$10 + m = \frac{5}{0.15} = 33.33$

$m = 33.33 - 10 = 23.33\,\text{kg}$

Match with options

The computed value is: $m \approx 23.3\,\text{kg}$

But this is not present in the options.

Check likely interpretation used in standard solution

In many such problems, the friction is taken due to the total load on the table including the original block and added weight, and if an additional block arrangement or figure implies a different effective pull, the answer may change. However, with the data provided as text only, the correct equilibrium equation is: $5g = 0.15(10+m)g$ Rightarrow m = 23.3,\text{kg}$$

So the stored answer $27.3\,\text{kg}$ cannot be obtained from the given textual information.

Conclusion

Based on the provided data, the minimum mass required is: $\boxed{23.3\,\text{kg}}$

Since this does not match any option, the problem statement likely depends on a missing figure or there is an error in the options/stored answer.