Answer:
(b) io(t) = -6e^(-3000t) mA, t > 0
Step-by-step explanation:
The inductor current at t=0 is 12 V/(2 kΩ) = 6 mA. When the switch is opened, that same current will continue to flow in the loop on the left. Its direction will continue to be down through the inductor, so in the negative direction with respect to the way io is defined.
This already tells you the correct answer choice: B.
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The circuit time constant is L/R = (2 H)/(6 kΩ) = 1/3000 seconds. The exponent of the exponential decay will be t divided by this, or -3000t, also in agreement with choice B.