What is the back voltage generated by this inductor in this practice problem?
The moment the switch opens, the top leg of the inductor rapidly falls negative to try and keep the current of 2A flowing. If the 4 ohms resistor were not present, a large negative voltage would appear at the top of the inductor and it would be large enough to cause the breakdown of air and a spark would form thus keeping current flowing out of the bottom of the inductor and into the very negative top.

But the 4 ohm resistor is there so no spark forms - there is a conventional discharge path for the current. So now the inductor uses the capacitor and 4ohm path for forcing current.

That initial 2A current flows through the 4 ohm resistor and creates a volt drop of 8 volts. The bottom end of the 4 ohm is attached to a capacitor charged at 12V so, the top of the 4 ohm becomes instantly 4 volts. 

Given that the 2A also flows through the 6 ohm resistor creating another volt drop of 12 volts means that the voltage at the top of the coil is 12 volts lower than 4 volts i.e. -8V.