Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the average power supplied by the emf, for the three wires, is given by
$$P_{\rm source}=P_1+P_2+P_3$$
$$P_{\rm source}=I_{\rm rms}\varepsilon_{\rm rms}+I_{\rm rms}\varepsilon_{\rm rms}+I_{\rm rms}\varepsilon_{\rm rms}$$
Hence,
$$I_{\rm rms}=\dfrac{P_{\rm source}}{3\varepsilon_{\rm rms}}\tag 1$$
Plug the known;
$$I_{\rm rms}=\dfrac{(450\times 10^6)}{3(120)}$$
$$I_{\rm rms}=\color{red}{\bf 1.25\times 10^6}\;\rm A$$
$$\color{blue}{\bf [b]}$$
Plugging the new data into (1),
$$I_{\rm rms}=\dfrac{(450\times 10^6)}{3(500\times 10^3)}$$
$$I_{\rm rms}=\color{red}{\bf 300}\;\rm A$$
$$\color{blue}{\bf [b]}$$
Any cable wouldn't be able to handle a current of 1.25 million A, it will melt, but it can handle the 300 A. That's why the step-up transformers are used here.
