Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the average power for an $RLC$ circuit is given by
$$P_{\rm source}=I_{\rm rms}\varepsilon_{\rm rms}\cos\phi$$
where $\cos\phi$ is the power factor.
$$\cos\phi=\dfrac{P_{\rm source}}{I_{\rm rms}\varepsilon_{\rm rms}}$$
Plug the given;
$$\cos\phi=\dfrac{ 800}{(8)(120)}=\color{red}{\bf 0.833}$$
$$\color{blue}{\bf [b]}$$
The rms resistor voltage is given by
$$V_{\rm rms}=\varepsilon_{\rm rms}\cos\phi$$
Plug the known;
$$V_{\rm rms}=(120)(0.833)=\color{red}{\bf100}\;\rm V$$
$$\color{blue}{\bf [c]}$$
The motor resistance is given by
$$R=\dfrac{V_{\rm rms}}{I_{\rm rms}}$$
Plug the known;
$$R=\dfrac{100}{8}=\color{red}{\bf 12.5}\;\rm \Omega$$
$$\color{blue}{\bf [d]}$$
When the power factor equals 1, it means that $\cos\phi=1\Rightarrow \phi=0^\circ$.
Recalling that
$$\tan\phi=\dfrac{X_L-X_C}{R}$$
So,
$$\tan0^\circ=\dfrac{X_L-X_C}{R}=0$$
Thus,
$$X_L=X_C\tag 1$$
Now we need to find the initial $\phi$,
$$\phi=\cos^{-1}(0.833)=\bf 33.56^\circ$$
Hence,
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]=33.56^\circ $$
Hence,
$$\tan(33.56^\circ)=\dfrac{X_L-X_C}{R}$$
So,
$$ [X_L-X_C]_i=R\tan(33.56^\circ)=(12.5)\tan(33.56^\circ)=\bf \bf 8.292\;\Omega$$
So,
$$X_L=X_C+8.292\tag 2$$
From (1) and (2), it is obvious that we need to increase the capacitive reactance by 8.292 $\Omega$.
So,
$$\Delta X_C=\dfrac{1}{2\pi f \Delta C}=8.292$$
Solving for $C$ to find the capacitance needed,
$$\Delta C=\dfrac{1}{2\pi f (8.292)}=\dfrac{1}{2\pi (60) (8.292)}$$
$$\Delta C=\color{red}{\bf 320}\;\rm \mu F$$
