Let the original center of mass remain fixed.

For the center of mass to stay at the same position, the net change in the numerator of $x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2}$ must be zero.

Suppose mass $m_1=3\,\text{kg}$ moves toward the center of mass by $2\,\text{cm}$. Its displacement relative to the original center of mass is $d_1 = 2\,\text{cm}.$

Let mass $m_2=2\,\text{kg}$ move toward the center of mass by $d_2$. Since the two masses are on opposite sides of the center of mass, moving both toward the center means their displacements are opposite in sign. Hence, for the center of mass to remain fixed, $m_1 d_1 = m_2 d_2.$

Substitute the values: $3 \times 2 = 2 \times d_2$ $6 = 2d_2$ $d_2 = 3\,\text{cm}.$

Therefore, the second particle must move $3\,\text{cm}$ toward the center of mass.