1. **State the problem:** We are given the rate of change of the area of a lamina as $$\frac{dA}{dt} = e^{-0.2t}$$ where $t$ is in seconds. The initial area at $t=0$ is 140 cm². We need to find the area after $\frac{1}{3}$ minutes (which is 20 seconds). 2. **Formula and explanation:** The area $A(t)$ can be found by integrating the rate of change: $$A(t) = A(0) + \int_0^t \frac{dA}{dt} dt = A(0) + \int_0^t e^{-0.2t} dt$$ 3. **Calculate the integral:** $$\int_0^{20} e^{-0.2t} dt = \left[-\frac{1}{0.2} e^{-0.2t} \right]_0^{20} = -5 \left(e^{-0.2 \times 20} - e^0 \right) = -5 (e^{-4} - 1) = 5(1 - e^{-4})$$ 4. **Find the area at $t=20$ seconds:** $$A(20) = 140 + 5(1 - e^{-4}) = 140 + 5 - 5e^{-4} = 145 - 5e^{-4}$$ 5. **Interpretation:** The area after $\frac{1}{3}$ minutes (20 seconds) is $145 - 5e^{-4}$ cm². **Answer:** (b) $145 - 5e^{-4}$