The effective current I in the given circuit at very high...

At very high frequencies, inductor acts as an open circuit and capacitor as short circuit.

So, capacitive reactance, {X}_{C} = \dfrac{1}{ωC} ≈ 0

Also, inductive reactance, {X}_{L} = ωL ≈ \infty

Thus, equivalent circuit can be redrawn as

Here, both 4Ω resistances are in parallel and together they are in series with1Ω and 2Ω resistances.

Equivalent impedance is Z = 1 + \dfrac{4 × 4}{4 + 4} + 2 = 5Ω

Effective current is I = \dfrac{220V}{5Ω} = 44A