What is the correct way to solve this DC RLC circuit?

I am studying for the FE Mechanical exam and am working my way through the electricity & magnetism section. Attached is a practice problem statement I found online and am having a bit of trouble working through.

My thought is that after the switch is closed the inductor will still act as a short while the capacitor will continue to act as an open circuit (with no current through that branch) since they are still connected to the voltage source. To that end I have redrawn a simplified circuit and applied KCL to solve for the "new" current through the 4K resistor. I end up with a result of 6.67mA. I understand that these exams often don't give the exact answer as a solution so my thought it that the correct answer would be 10mA. However, I am not confident that I have solved this question correctly.

Can anyone review this work and let me know of any missteps that I may have made? Any feedback is appreciated. This is my first post on the site so if something is unclear please let me know! Thank you.

The inductor will oppose the change in steady state, hence, at the instant the switch is closed, the current through the inductor will be the same as the steady state value. Therefore $$\\small I= \:... \$$
• Yes. You can safely ignore all the components to the left of the inductor - put them in a 'black box' if you wish. Your analysis is now correct, $\small I=5\:\text{mA}$