Answer
$\bf Lags.$
Work Step by Step
This is the opposite of the previous problem in which we found that $\phi\lt 0$ and this gave us $\omega\lt \omega_0$.
We are given that
$$\omega\gt \omega_0$$
where $\omega_0=1/\sqrt{LC}$, so
$$\omega\gt \sqrt{\dfrac{1}{LC} }$$
$$\omega^2\gt \dfrac{1}{LC} \tag 1 $$
Recalling that:
$X_{\rm C}=1/\omega C$, so $C=1/\omega X_{\rm C}$, and
$X_{\rm L}=\omega L$, so $L=X_{\rm L}/\omega$
Plug into (1),
$$\omega^2\gt \dfrac{1}{LC} $$
$$\omega^2\gt \dfrac{1}{(X_{\rm L}/\omega)(1/\omega X_{\rm C})} $$
$$ \color{red}{\bf\not} \omega^2\gt \dfrac{ \color{red}{\bf\not} \omega^2 X_{\rm C}}{ X_{\rm L} } $$
$$1\gt \dfrac{ X_{\rm C}}{ X_{\rm L} } $$
Therefore,
$$ X_{\rm L}\gt X_{\rm C}$$
$$ X_{\rm L}-X_{\rm C}\gt 1 $$
which means that $\boxed{\phi\gt 0}$ where
$$\phi=\tan^{-1}\left[ \dfrac{X_{\rm L}-X_{\rm C}}{R} \right]$$
This means that the current lags the emf.
