Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the change in potential energy of charge $Q$ is given by
$$\Delta U=Q\Delta V$$
And since the charge $Q$ that passes through the wire is experiencing a potential difference of $\Delta V_{bat}$
$$\boxed{\Delta U=Q\Delta V_{bat}}$$
$$\color{blue}{\bf [b]}$$
This energy is the cause of the warming up of the wire, so this energy is transferred to heat energy in the wire.
$$\color{blue}{\bf [c]}$$
We know that the power is given by
$$P=\dfrac{E}{t}$$
and in our case, $E$ is $\Delta U$, so
$$P=\dfrac{\Delta U}{t}= \dfrac{Q\Delta V_{bat}}{t}$$
Recalling that $I=Q/t$, hence
$$\boxed{P=I\Delta V_{bat} }$$
$$\color{blue}{\bf [d]}$$
Plug the given into the last boxed formula above,
$$P=I\Delta V_{bat}=(1.2)(1.5)$$
$$P=\color{red}{\bf 1.8}\;\rm W$$
