Voltage Amplifier?

Relevant wiki: Impedance

The impedance of the circuit $Z = {X_L} - {X_C}$
Where, $X_L$ and $X_C$ are the inductive and the capacitive reactances respectively.
Also, we have, ${X_L} = 2{X_c}$
Therefore,
$Z = \frac{{{X_L}}}{2}$
The peak value of current in the circuit will be
${i_0} = \frac{{{V_0}}}{Z}$
The peak value of voltage across the inductor will be
${V_{{L_0}}} = {i_0}{X_L} = {V_0}\frac{{{X_L}}}{Z} = {V_0}2$
Hence, $\frac{{{V_{{L_0}}}}}{{{V_0}}} = 2$

Bonus Solution: Let us draw the phasor diagram We can see clearly from the phasor diagrams that the voltage across inductor and the source voltage are in the same phase.

Let the AC voltage be represented as $V(t) = V_0 \cdot sin(\omega t).$ Given that the inductive impedance is twice that of the capacitive impedance, we have:

$|j\omega L| = 2 \cdot |\frac{1}{j\omega C}| \Rightarrow \omega L = \frac{2}{\omega C} \Rightarrow \omega^{2}LC = 2.$

The voltage across the inductor is determined via a KVL voltage-divider:

$V_L (t) = V(t) \cdot \frac{|Z_L|}{|Z_L + Z_C|} = V(t) \cdot \frac{\omega L}{\omega L - \frac{1}{\omega C}} = V(t) \cdot \frac{\omega^{2} LC}{\omega^{2} LC - 1} = V(t) \cdot \frac{2}{2 - 1} = 2V(t).$

Hence, at AC steady-state: $\frac{V_L (t)}{V(t)} = \frac{2 \cdot V(t)}{V(t)} = \boxed{2:1}.$