You would only be covering the bottom of the resistor with the tape. The resistor is in parallel with the inductor, so the vast majority of current to the drivers will be delivered through the copper windings, not the resistor. The resistor will only see portions of the signal that the inductor begins to resist, which would be high frequencies that the ribbon itself will see most of. So your current calculations need to consider that the resistor is also in parallel with the rest of the network and ribbon. It also seems a bit extreme to expect 9 amps across that resistor - that would take 189Vrms (16 ohm resistor + 7 ohm driver network) which only a 4500W amplifier could possibly do.