BJT common-emitter analysis
To find the \$R_B\$ and \$R_C\$ values you can use this approach: 

Because of the fact, the lower side of a \$R_B\$ resistor is at 0V and you want 0.2V at the top we can find this resistor using only the Ohms law. 

$$R_B = \frac{0.2V}{I_B} = \frac{0.2V}{\frac{I_{Eq}}{\beta + 1}} = \frac{0.2V}{9.88\mu A} \approx 20k\Omega$$  

And for the \$R_C\$, you want -3V at the collector wich means that the voltage drop across \$R_C\$ resistor is equal to: 
$$V_{R_C} = (-3V) - (-5V) = 2V $$  what about the current?  

We know that: 

$$I_E = I_B + I_C = I_B + \beta\cdot I_B = (\beta +1)I_B$$ and 

$$\frac{I_C}{I_E} =\frac{I_C}{I_B +I_C}= \frac{\beta \cdot I_B}{I_B +\beta \cdot I_B}=  \frac{\beta}{\beta + 1}$$

$$I_C = I_E \frac{\beta}{\beta + 1} = 0.8\textrm{mA} \cdot \frac{80}{81} \approx 0.79\textrm{mA} $$  Therefore: 

$$R_C = \frac{2V}{0.79\textrm{mA}} = 2.53k\Omega$$ 

Is this what you want?