Can I use unregulated power supply for LCD display backlight LED?
If you're going to use a series resistor then it has to drop 24 - 6.4 = 17.6 V. If you want 20 mA then \$ R = \frac V I = \frac {17.6}{20m} = 0.88\ \mathrm{k\Omega} \$ or 820 Ω for the nearest E12 value. Recalculating for that gives us \$ I = \frac V R = \frac {17.6}{820} = 21.5 \ \text{mA}\$.
The voltage will remain fairly constant across the LEDs over a wide range of currents so we can just calculate the change in current caused by the resistor voltage varying. The 2 V ripple will give a 2/17.6 = 11% ripple in the current.
The fact that the LEDs are on bright in the first place, the fact that the ripple will be mains or double mains frequency (depending on the rectifier) and the persistence of human vision will make it most unlikely that you will observe any flicker.
Power dissipated in the resistor will be given by \$ P = I^2 R = 0.0215^2 \times 820 = 0.38 \ \text W \$. Make up the resistance using a few series or parallel with twice the required wattage total so that it all runs cool.
I chose 820 Ω. You could go for 1 kΩ and reduce the current a little.