A coil of inductance 8.4 mH and resistance 6 Ω is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the time

Peak current in the circuits : $i_0=\frac{12}{6}=2A$  . As per the situation given the circuit is a decaying series LR circuit.
Current decreases from 2A to 1A i.e., becomes half in time:
$t=0.693\frac{L}{R}=0.693\times \frac{8.4\times {10}^{-3}}{6}=1\mathit{ }milli\mathit{ }sec$