As depicted in the figure, a uniform rod of length 30 cm and mass...

Total mass of the rod, $M = 3.0 \text{ kg}$
Total length of the rod, $L = 30 \text{ cm}$

The net force acting on the rod is:
$F_{\text{net}} = 32 - 20 = 12 \text{ N}$ towards the right.

The acceleration of the rod is:
$a = \dfrac{F_{\text{net}}}{M} = \dfrac{12}{3.0} = 4 \text{ m/s}^2$

Since the rod is uniform, the mass of the $10 \text{ cm}$ part is:
$m_1 = \dfrac{10}{30} \times 3.0 = 1.0 \text{ kg}$

Let $F$ be the force exerted by the $20 \text{ cm}$ part on the $10 \text{ cm}$ part. The forces acting on the $10 \text{ cm}$ part in the horizontal direction are $20 \text{ N}$ to the left and $F$ to the right.

Applying Newton's second law to the $10 \text{ cm}$ part:
$F - 20 = m_1 a$
$F - 20 = 1.0 \times 4$
$F = 24 \text{ N}$