How to calculate the base-current of this transistor?
It depends on how much precision you think you need....

The simple approach is to simply know that Vbe for a silicon transistor in conduction is ~0.6V or so, and that this voltage must thus appear across R1, thus I(r1) must be ~ 6mA, V(r2) = 20 - 0.6 = 19.4V, I (r2) = 19.4/2.7k = ~7.2mA, so Ib = 7.2 - 6 = 1.2mA (I leave the signs to you as that depends on how you define the current loops).

The more precise approach (and if you need this, you are probably doing it wrong) is to use the Ebbers-Moll equation which relates Ib to Vbe and temperature (The base-emitter junction is basically a silicon diode), but that turns into a mess of maths because the Vbe ->Ib relation is exponential, and you almost never need this (Unless you are Jim Williams). 

R3 provides the base current for the second transistor (And by the way this circuit will destroy itself as you power it up, because until the supply voltage becomes sufficient to turn the first transistor on, there is little to limit the current in D1 and the second transistor except perhaps for the value of R3, but that depends on the HFE of the second transistor that is not a well controlled parameter.

Regards, Dan.