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.
So, the current leaving the station is given by
$$I_{\rm rms}=\dfrac{P_{\rm source}}{\varepsilon_{\rm rms}\cos\phi}$$
Plug the given;
$$I_{\rm rms}=\dfrac{(6\times 10^6)}{(15,000)(0.90)}$$
$$I_{\rm rms}=\color{red}{\bf 444}\;\rm A$$
$$\color{blue}{\bf [b]}$$
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 $\phi$,
$$\phi=\cos^{-1}(0.833)=\bf 25.842^\circ $$
So, from the geometry of the figure below, $\sin\phi=\dfrac{X_L-X_C}{Z}$,
Hence,
$$\sin\phi=\dfrac{X_L-X_C}{Z} $$
where $Z=\varepsilon_{\rm rms}/I_{\rm rms}$
$$\sin\phi=\dfrac{(X_L-X_C)I_{\rm rms}}{\varepsilon_{\rm rms}} $$
Thus,
$$X_L-X_C= \dfrac{\varepsilon_{\rm rms} \sin\phi}{I_{\rm rms}}$$
Plug the known;
$$X_L-X_C= \dfrac{(15000)\sin(25.842^\circ)}{(444.44)}=14.7$$
Therefore,
$$X_L=X_C+14.7\tag 2$$
From (1) and (2), it is obvious that we need to increase the capacitive reactance by 14.7 $\Omega$.
So,
$$X_C=\dfrac{1}{2\pi f C}=14.7$$
$$C=\dfrac{1}{2\pi (60) (14.7)}=\color{red}{\bf 180}\;\rm \mu F$$
$$\color{blue}{\bf [c]}$$
The power delivered by the station is then given by
$$P_{\rm source}'=\dfrac{\varepsilon_{\rm rms}^2}{R}$$
where, from (2), $ \tan(25.842^\circ)=\dfrac{X_L-X_C}{R}=\dfrac{14.7}{R}$, so $R=\dfrac{14.7}{ \tan(25.842^\circ)}$
$$P_{\rm source}'=\dfrac{\varepsilon_{\rm rms}^2\tan(25.842^\circ)}{(14.7)}=\dfrac{(15000)^2\tan(25.842^\circ)}{(14.7)}$$
$$P_{\rm source}'=\color{red}{\bf 7.41}\;\rm MW$$
