Center of gravity

Initial position of center of mass of the system (both men and the boat too) from the center of the boat = $\frac{50\times(-2) + 40\times 0 + 60\times 2}{50 + 40 + 60}m = \frac{2}{15}m$

And finally the center of mass of the system will be at the center of the boat.

Since no external forces are acting on the system, so center of mass of the system should not change its position throughout this event.

But there is a gap of $\frac{2}{15}m$ between initial and final distance of center of mass of the system from the center of the boat.

This means that the center of the boat must move $\frac{2}{15}m$ i.e. $13.33 cm$ through the water so that the center of mass of the system do not change its position.

Interesting question! The answer comes from the Law of Conservation of Momentum.

First consider the $m_1 =60$ kg man walks \frac{b}{a^2} to the center of the boat with a velocity $v_1$ . Since the initail momentum before the man walking is $0$ , the total mass $M$ kg of the boat and the two men move in the opposite direction with velocity $v_3$ such that: $m_1v_1=Mv_3$
And since both the man and the total mass move with same time period, the distances $s_1=2$ m and $s_3$ m are proportional to their velocities, therefore, $m_1s_1=Ms_3$ $\Rightarrow s_3 = \cfrac{m_1}{M}s_1 = \cfrac{60}{50+60+40}\times 2=0.8m$ Similarly, when the 50-kg man walks to the boat center: $s_4=\cfrac{50}{150}\times 2 = 0.667m$ Since $s_3$ and $s_4$ are in opposite directions, the resultant distance $s$ the boat moves is: $s=s_3-s_4=0.8-0.667=0.133m\approx \boxed{13.0cm}$

My solution inspired by @Arjen Vreugdenhil and @Aryan Sanghi .

$\text{Initially,}$

$\dfrac{50\times 0 + 40 \times 2 +60\times 4}{50+40+60}=\dfrac{80+240}{150}=\dfrac{32}{15} \text{m}$

$\text{After moving,}$

$\dfrac{0\times 0 + (50+40+60) \times 2 +60\times 0}{50+40+60}=2=\dfrac{30}{15} \text{m}$

$\text{Change}$

$\dfrac{32}{15}\text{m}-\dfrac{30}{15} \text{m}=\dfrac{2}{15} \text{m}=\dfrac{40}{3} \text{cm} \approx \boxed{13 \text{cm}}$