BJT average power dissipation
In short. 

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2\text{mA} \times 6.97\text{V} \approx 13.94 \text{mW}$$

And the AC component of a collector current is equal to 

$$i_c = g_m v_{\pi} = 1/13\Omega \times  18 \text{mV} \approx 1.38 \text{mA}$$

Therefore power dissipated in the transistor is: 

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94\text{mW} - \frac{1.38\text{mA}^2 \times 1\text{k}\Omega}{2} \approx 13\text{mW}$$