Answer
${\bf43.5}\;\rm A$
Work Step by Step
We know, in $RLC$ circuits, that the energy dissipated by the resistor is $$P_R=I_{\rm rms}V_{\rm rms} $$
where
$$V_{\rm rms}=\varepsilon_0 \cos\phi\tag 1$$
where $\phi$ is the phase difference between the current and the emf.
Now we can use Ohm's law,
$$V=IR$$
Plug into (1)
$$I_{\rm rms}R=\varepsilon_0 \cos\phi $$
$$R=\dfrac{\varepsilon_0 \cos\phi}{I_{\rm rms}}$$
Plug the given;
$$R=\dfrac{(120)(0.87)}{(2.4)}=\color{red}{\bf43.5}\;\rm A$$
