Answer
$\bf Less\;than.$
Work Step by Step
It is obvious, from the given graph, that the current leads the emf.
This means that $\phi\lt 0$ where $\phi$ is given by
$$\phi=\tan^{-1}\left[ \dfrac{X_{\rm L}-X_{\rm C}}{R} \right]$$
So,
$$ \tan^{-1}\left[ \dfrac{X_{\rm L}-X_{\rm C}}{R} \right]\lt 0$$
which means that
$$ X_{\rm L}-X_{\rm C} \lt 0\tag 1$$
Recalling that:
$X_{\rm C}=1/\omega C$, and
$X_{\rm L}=\omega L$,
Plug those into (1),
$$ \omega L-\dfrac{1}{\omega C} \lt 0 $$
$$\dfrac{ \omega^2 LC-1}{\omega C} \lt 0 $$
$$ \omega^2 LC-1 \lt 0 $$
$$ \omega^2 LC \lt 1 $$
$$ \omega^2 \lt \dfrac{1}{LC} $$
$$ \omega \lt \sqrt{\dfrac{1}{LC} }$$
where $\omega_0=1/\sqrt{LC}$, so
$$\boxed{ \omega \lt \omega_0 }$$
