A coil of inductance 1.8 H carries a steady 0.80 A…

Step 1. The time constant. $\tau = \dfrac{L}{R} = \dfrac{1.8\ \text{H}}{45\ \Omega} = 0.040\ \text{s} = 40\ \text{ms}$.

Step 2. The half-time. Set $e^{-t/\tau} = \dfrac{1}{2}$. We want $t$, so $t = \tau\ln 2 = 40\ \text{ms} \times 0.693 = 27.7\ \text{ms}$.

Step 3. The heat. $W = \tfrac{1}{2}LI_0^2 = \tfrac{1}{2} \times 1.8\ \text{H} \times (0.80\ \text{A})^2 = 0.576\ \text{J}$.