= Solution
Initially the open switch leaves a series capacitor-resistor-inductor circuit driven by the battery. After a long time the <electric current> is zero and the <capacitor voltage> is $V_0$. Closing the switch bypasses and, under the stated idealization, instantaneously discharges the capacitor. The inductor current remains continuous at zero.
The remaining <RL circuit> obeys $L\dot I+RI=V_0$, so
$$
I(t)=\frac{V_0}{R}\left(1-e^{-Rt/L}\right),\qquad\boxed{I(0.005\,\mathrm s)=2(1-e^{-1.875})\simeq1.69\,\mathrm A.}
$$
The capacitor's impulsive discharge through the ideal short is distinct from the finite <electric current> through the coil. A finite switch resistance would resolve that separate fast transient.
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