Using Thévenin's Theorem / Voltage Across Independent Current Sources
1) 2mA goes through 1k, combines with 9mA, then 2+9 = 11mA goes through 6k.
2) Thus the voltage drop across 1k is 2m*1k = 2V
3) Voltage drop across 6k is 11m*6k = 66V
4) Just add up the voltages:  5V + 66V + 2V = 73V is the open circuit voltage across 4k resistor.