A coil has 50 turns, and each turn has an area of 40 cm²…

Step 1. The emf. The flux linkage is $N\Phi = NAB_0e^{-t/\tau}$, with $NAB_0 = 50 \times 4.0 \times 10^{-3}\ \text{m}^2 \times 0.60\ \text{T} = 0.12$ Wb. The size of the emf is $\varepsilon = \dfrac{NAB_0}{\tau}\,e^{-t/\tau}$. At $t = 0$: $\varepsilon = \dfrac{0.12\ \text{Wb}}{0.20\ \text{s}} = 0.60$ V. At $t = \tau$ the factor $e^{-1} = 0.368$ gives $0.221$ V.

Step 2. The charge. The flux linkage falls from $0.12$ Wb to zero, so $Q = \dfrac{N\Delta\Phi}{R} = \dfrac{0.12\ \text{Wb}}{3.0\ \Omega} = 0.040\ \text{C}$, that is $40$ mC.

Step 3. The heat. $H = \int_0^\infty \dfrac{\varepsilon^2}{R}\,dt = \dfrac{(0.60\ \text{V})^2}{3.0\ \Omega}\int_0^\infty e^{-2t/\tau}\,dt = \dfrac{0.36\ \text{V}^2}{3.0\ \Omega} \times \dfrac{0.20\ \text{s}}{2} = 0.012\ \text{J}$, that is $12$ mJ.