AC makes things complicated.
Here is the way to calculate it with DC. (The peak of 5 VAC is 7.07 V DC.)
Subtract the diode drop. (Common practice is to use 0.6V, although a typical range is 0.4 to 0.8 V for 1N400x series.)
Assume the resistor is by itself. Calculate current through it.
( 7.07 - 0.6 ) / 220 = 29.4 mA
The identical current is going through the diode.
Therefore diode resistance is V / A.
0.6 / .0294 = 20.4 ohms
Its dynamic resistance will vary throughout a sine cycle. Between 20.4 ohms and infinity. That is through one half of the cycle. During the other half it is not conducting.
If you desire accuracy, then do further iterations. Recalculate the diode drop between iterations.