Key idea at resonance in an RLC circuit

At resonance, the inductive and capacitive reactances cancel: $X_L = X_C$ so the net reactance of the LC part becomes zero.

Interpretation of the given circuit

Since the question specifically asks for the current through the $55\,\Omega$ resistance at resonance, the standard configuration for such a question is that the $55\,\Omega$ resistor is connected across the branch whose effective voltage becomes zero at resonance.

At resonance, the potential difference across that branch is zero, hence the current through the $55\,\Omega$ resistor is: $I = \frac{V}{R} = \frac{0}{55} = 0$

Current through the $55\,\Omega$ resistor

Therefore, $I_{55\Omega} = 0\text{ A}$

Final integer answer

$\boxed{0}$