Here we are faced with at least three different quantities:

Efficiency in its usual meaning is the ratio of the actual output power to the actual input power. With ideal transformer and ideal diode, the efficiency would be 1 or 100 % because neither the transformer nor the diode turns any electrical energy into heat or another form of energy, so the principle of conservation of energy demands that in the stationary state output power = input power.

Conversion ratio compares the DC component of the output power (which is less than the actual output power because the power spectrum at the output contains also the AC frequency plus a considerable amount of harmonics) with the actual input power. Because input current flows only during the positive half waves of the voltage, the input power is 1/2 * Vpeak/sqrt(2) * Ipeak/sqrt(2). The DC component of the output power equals the mean of the voltage (not the RMS of the voltage) times the mean value of the current. With ideal transformer and ideal diode, the conversion ratio is about 0.405 resp. 40.5 %.

You are introducing a third quantity (for the sake of clarity, I wouldn't call it efficiency) which is, as far as I can see, the ratio of the DC component of the output power to the power received by RL, if it would be connected directly to the AC voltage. Since this "input power" equals Vpeak/sqrt(2) * Ipeak/sqrt(2), your input power is twice the actual input power as used in conversion ratio, so your limit for ideal transformer and ideal diode is only half the limit for the conversion ratio.

As to your first question, since there are multimeters which display the rms values you can use them to measure the actual input power and the actual output power (provided the bandwidth is sufficient). In order to calculate the DC part of the output power, I would measure voltage and current with a digitizing scope (using suitable probes, of course), transfer the results to a computer, and calculate the DC power.

To your second question, it depends on the problem you are trying to solve. At the moment, I cannot think of a real application which requires calculating your third quantity, but then I guess it's just as hard to imagine a real application which requires calculating the conversion ratio: Either RL is just a resistor then the power distribution in the output spectrum is of no consequence; or RL represents some kind of circuit which requires a DC supply, then we would add a capacitor parallel to RL which would render the second as well as the third quantity mentioned above useless. A third possibility would be to use the rectifier as an AM demodulator; but here, too, it would be common to add a capacitor.

As an afterthought, if RL would be replaced by an inductor and a resistor in series, and if the cut off frequency of this low pass filter would be sufficiently low to virtually suppress all spectral components in the output current except for the DC part, and if a second diode would be placed parallel to the LR combination, cathode connected to the cathode of the first diode, then we had a case where the DC component of the voltage behind the first diode really matters because it would equal the voltage at the resistor (with ideal components).

But unfortunately the actual currents would have no resemblance, neither to the currents used in the conversion ratio nor to your currents.