The net change in internal energy of a system can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

Given that the system undergoes four processes with heat transfers of 0, 8, 6, and -4 units, respectively, we can calculate the net change in internal energy as follows:

ΔU = Q1 + Q2 + Q3 + Q4 - W

where Q1, Q2, Q3, and Q4 are the heat transfers in each process, and W is the work done by the system.

Since the work done by the system is not given, we cannot calculate the net change in internal energy directly. However, we can use the fact that the net change in internal energy over a complete cycle is zero, since the system returns to its initial state. Therefore, we have:

ΔUcycle = 0 = Q1 + Q2 + Q3 + Q4 - Wcycle

where Wcycle is the net work done by the system over the cycle.

Since Wcycle is not given, we cannot calculate it directly. However, we can use the fact that the net work done by the system over a complete cycle is equal to the area enclosed by the cycle on a PV diagram. Therefore, if we plot the four processes on a PV diagram, we can calculate the area enclosed by the cycle and use it to find the net work done by the system.

Assuming that the system is ideal and undergoes a cyclic process, the area enclosed by the cycle on a PV diagram is a rectangle with sides equal to the initial and final volumes and pressures of the cycle. Therefore, we can calculate the net work done by the system as follows:

Wcycle = (P2 - P1)(V2 - V1)

where P1, V1, P2, and V2 are the initial and final pressures and volumes of the cycle.

Since the pressures and volumes are not given, we cannot calculate the net work done by the system directly. However, we can use the fact that the net work done by the system over a complete cycle is equal to the difference between the heat added to the system and the heat rejected by the system. Therefore, we have:

Wcycle = Q1 + Q2 + Q3 + Q4

Substituting this expression for Wcycle in the equation for ΔUcycle, we get:

ΔUcycle = 0 = Q1 + Q2 + Q3 + Q4 - Q1 - Q2 - Q3 - Q4

Simplifying, we get:

ΔUcycle = 0

Therefore, the net change in internal energy of the system over the complete cycle is zero. Since the system returns to its initial state, the net change in internal energy over any part of the cycle, including the four individual processes, is also zero. Therefore, the correct answer is option B, zero.