A battery of 12 V and negligible internal resistance is joined to a resistor of 6.0 Ω…

Step 1. Just after closing. The coil's current is zero, so all the current goes through $R_2$. The circuit is $R_1$ and $R_2$ in series: $I = \dfrac{12\ \text{V}}{6.0\ \Omega + 12\ \Omega} = 0.667\ \text{A}$. The voltage across the coil is then $0.667\ \text{A} \times 12\ \Omega = 8.0\ \text{V}$.

Step 2. After a long time. The coil carries a steady current and has no voltage across it. It acts like a wire that shorts out $R_2$. So $I = \dfrac{12\ \text{V}}{6.0\ \Omega} = 2.0\ \text{A}$, and all of it goes through the coil.

Step 3. The time constant. The coil sees the two resistors in parallel, because the battery has no resistance. $R_{eff} = \dfrac{6.0 \times 12}{6.0 + 12} = 4.0\ \Omega$. So $\tau = \dfrac{L}{R_{eff}} = \dfrac{0.80\ \text{H}}{4.0\ \Omega} = 0.20\ \text{s}$.