In reality, this question is pretty much impossible to answer as different motors draw different amounts of current at various loads.
I think the crux of your question is, what's the difference in current draw between a 1-phase and a 3-phase motor, all else being equal.
Since we're talking about AC here, voltages and currents are expressed in terms of their RMS values (root mean square). That's the equivalent DC (constant) voltage or current that would be required to produce the same amount of power as the AC signal. It is calculated as the square root of the average value of the square of the signal.
For a single-phase sinusoidal signal, we have:
V(t) = Vpeak*sin(wt), where w is the angular frequency (2pi*60 Hz in the US).
V^2(t) = Vpeak^2*sin^2(wt) = Vpeak^2/2*(sin(2wt)+1)
Since the average value of a sinusoid is 0, we have:
V^2(t)_ave = Vpeak^2/2
Finally, the RMS value is the square root of this:
Vrms = sqrt(V^2(t)_ave) = sqrt(Vpeak^2/2) = Vpeak * sqrt(2)/2
This is a very well-known formula that you've probably seen many times before.
Now let's consider a 3-phase voltage. In this case, the RMS value of the voltage is the square root of the average value of the sum of
the squares of the voltages, since each is contributing to the total delivered power.
V1(t) = Vpeak*sin(wt)
V2(t) = Vpeak*sin(wt-2pi/3)
V3(t) = vpeak*sin(wt+2pi/3)
The +/- 2pi/3 terms represent the 120 degree phase shift among the signals.
V^2 = V1^2 + V2^2 + V3^2
Substituting the formulas for V1, V2, and V3 from just above, and using the trig identity that sin(x+y)=sin(x)cox(y)+sin(y)cos(x), and knowing that cos(2pi/3)=cos(-2pi/3)=-1/2, we get:
V^2 = Vpeak^2 * 3/2
Since the result is actually a constant, not depending on t, finding the average is trivial; it IS the average.
Thus RMS voltage for a 3-phase signal is:
Vrms_3phase = Vpeak * sqrt(6) / 2
Compared to the single-phase case we get:
Vrms_3phase = sqrt(3) * Vrms_1phase
With an RMS value sqrt(3) times larger than single-phase, a 3-phase source will therefore deliver sqrt(3) times as much power to a given load. A corrolary to this is that at a constant voltage, a single-phase source requires sqrt(3) times as much current to deliver the same amount of power as a three-phase source.
In the case of a 2HP motor, therefore, the current drawn in the single-phase case should be about 1.732 times higher than for the 3-phase case. You can see that factor reflected in the motor current ratings shown here: The Panel Shop: PanelFacts - Motor Current Ratings
In the case of a 100% efficient motor, 2HP of mechanical power would be generated by exactly 2HP of electrical power. Converting to watts, we get:
2HP * 746 watts/HP = 1492 watts
At 220V, this means the 1-phase motor would be drawing:
I_1phase = 1492/220 = 6.8 amps,
I_3phase = 6.8/sqrt(3) = 3.9 amps.
Finally, let me reiterate that just because that's what these fictitious 100% efficient motors would theoretically draw, that is NOT a realistic current rating for a real motor. Actual current draw varies greatly among motors, even at the same power rating, and will also vary greatly with rotational speed, torque, etc. For example, as you probably know, current draw at startup is MUCH higher than at full speed when the motor is generating significant back-EMF. Hence, on the link above you see a panel rating of 24 amps for a 2HP single-phase motor at 220V, nearly 4 times higher than the theoretical current draw!