To find the currentI (in A) through the 6 Ω resistor, we need to analyze the given circuit using Kirchhoff's Voltage Law (KVL) and Current Law (KCL).

Step-by-step solution:

1. The circuit includes a current source of 20 A connected in parallel with a series combination of a 2 Ω resistor and a voltage source of 60 V on the left side, and a 3 Ω resistor and a voltage source of 30 V on the right side.
2. Apply KCL at the top node where the current source, 2 Ω, 6 Ω, and 3 Ω resistors meet. The current entering the node from the current source is balanced by the current flowing through the resistors:
   • The total current entering the node from the current source is 20 A.
   • This current splits into parts through the 2 Ω, 6 Ω, and 3 Ω resistors.
3. Apply KVL in the loop with the 2 Ω resistor and the left voltage source:
4. Apply KVL in the loop with the 3 Ω resistor and the right voltage source:
5. For the 6 Ω resistor:
6. Using KCL:
   • Solve the above equations considering I_1, I_3, and V to find the current I (in A).
7. Using equations, we get:

Thus, the current through the 6 Ω resistor is 10 A. The correct answer is 10.