Calculating capacitor capacitance between collector and emitter in radio circuits
The circuit in the link won't work very effectively because Q1 is biased incorrectly but, ignoring that little problem, what you have (and ignoring the superfluous stuff) is a simple oscillator: -

Basically it's a common-base oscillator that performs frequency modulation when the base bias point is superimposed with audio. This has the effect of altering the Miller capacitor (internally between base and collector) and therefore modifying the tank oscillating frequency.
I've estimated the DC bias for this final stage given that the previous stage base bias resistors are incorrect. So, what you get is an internal \$r_E\$ for the BJT (Q2) of about 2 ohms.
For AC signal analysis, \$r_E\$ is regarded as connected to a voltage source (the &quot;true&quot; emitter) and therefore it can be regarded as grounded for signals. This largely means that R6 can be ignored (because it would be in parallel with \$r_E\$ for AC signal analysis).
The value of \$r_E\$ is important because C5 can de-tune the L1/C6 resonant peak and, any significant detuning will lower the Q-factor of the resonant tank. This is because \$r_E\$ is in series with C5. You don't want \$r_E\$ to be a significant part of the tuning hence, C5 is kept approximately ten times lower than C6. It's a bit of a rule of thumb of course.

My question is that, how to know/calculate farads of that capacitor?

For this particular circuit it's a 10:1 rule of thumb. For a more conventional and reliable oscillator like a Colpitts type, the rules are different. Horses for courses.