How to calculate the power dissipation in a transistor?

Power isn't "across" something. Power is the voltage across something times the current going through it. Since the small amount of current going into the base is irrelevant in power dissipation, calculate the C-E voltage and the collector current. The power dissipated by the transistor will be the product of those two.

Let's take a quick stab at this making some simplifying assumptions. We'll say the gain is infinite and the B-E drop is 700 mV. The R1-R2 divider sets the base at 1.6 V, which means the emitter is at 900 mV. R4 therefore sets the E and C current to 900 µA. The worst case power dissipation in Q1 is when R3 is 0 so that the collector is at 20 V. With 19.1 V accross the transistor and 900 µA through it, it is dissipating 17 mW. That's not enough to notice the extra warmth when putting your finger on it, even with a small case like SOT-23.

Power is the rate at which energy is being converted into some other energy. Electrical power is the product of voltage and current:

$$ P = VI $$

Usually we are converting electrical energy into heat, and we care about power because we don't want to melt our components.

It doesn't matter if you want to calculate the power in a resistor, transistor, circuit, or waffle, power is still the product of voltage and current.

Since a BJT is a three-terminal device, each of which may have a different current and voltage, for the purposes of power calculation it helps to consider the transistor as two parts. Some current enters the base, and leaves the emitter, through some voltage \$V_{BE}\$. Some other current enters the collector and leaves the emitter through some voltage \$V_{CE}\$. The total power in the transistor is the sum of these two:

$$ P = V_{BE}I_B + V_{CE}I_C $$

Since the goal of using a transistor is usually to amplify, the collector current will be much larger than the base current, and the base current will be small, small enough to be neglected. So, \$I_B \ll I_C\$ and the power in the transistor can be simplified to:

$$ P \approx V_{CE} I_{C} $$