A flat copper disc has a radius of 3.5 cm and a thickness of 1.5 mm…

Step 1. The emf at the rim. $\varepsilon = \pi a^2\dfrac{dB}{dt} = \pi \times (0.035\ \text{m})^2 \times 2.4\ \text{T/s} = 9.24 \times 10^{-3}\ \text{V}$. That is $9.24$ mV.

Step 2. The heat. $P = \dfrac{\pi t a^4}{8\rho}\left(\dfrac{dB}{dt}\right)^2 = \dfrac{\pi \times 1.5 \times 10^{-3}\ \text{m} \times (0.035\ \text{m})^4 \times (2.4\ \text{T/s})^2}{8 \times 1.7 \times 10^{-8}\ \Omega\,\text{m}} = 0.2995\ \text{W}$. That is $0.300$ W.

Step 3. The larger disc. The heat goes as $a^4$. So $P = 2^4 \times 0.2995\ \text{W} = 4.79\ \text{W}$.