| Not exactly. The diode current is feeding into a filter capacitor. During some or much of the cycle, the current is cut off because the diode ouput voltage is less than the capacitor voltage. So, the current waveform will not be a rectified sinewave - it will be sort of a truncated sinewave, so 0.707 isn't the correct factor. Even most Fluke "true RMS" meters won't give the correct current, because the current waveform isn't symmetrical about the zero axis. But the RMS current is only important for calculating the power rating of the resistor. But for that, estimate it or measure it anyway you want, and double for the resistor rating.
It's the B+ voltage that you are trying to match. For that, guess at the resistor ohms, and go by trial and error.
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