Consider the fig. (b) to solve this question,

$$ \begin{aligned} I_E & =I_C+I_B \text { and } I_C=\beta I_B \quad\text{.... (i)}\\ I_C R_C+V_{C E}+I_E R_E & =V_{C C} \quad\text{.... (ii)}\\ R I_B+V_{B E}+I_E R_E & =V_{C C} \quad\text{.... (iii)}\\ \because\quad I_E & \approx I_C=\beta I_B \end{aligned}$$

From Eq. (iii),

$$\left(R+\beta R_E\right) I_B=V_{C C}-V_{B E}$$

$$\begin{aligned} \Rightarrow \quad I_B & =\frac{V_{\mathrm{CC}}-V_{\mathrm{BE}}}{R+\beta \cdot R_E} \\ & =\frac{12-0.5}{80+1.2 \times 100}=\frac{11.5}{200} \mathrm{~mA} \end{aligned}$$

$$\begin{aligned} &\text { From Eq. (ii), }\\ &\begin{aligned} \left(R_C+R_E\right) & =\frac{V_{\mathrm{CE}}-V_{\mathrm{BE}}}{I_C}=\frac{V_{\mathrm{CC}}-V_{\mathrm{CE}}}{\beta I_B} \quad\left(\because I_C=\beta I_B\right)\\ \left(R_C+R_E\right) & =\frac{2}{11.5}(12-3) \mathrm{k} \Omega=1.56 \mathrm{k} \Omega \\ R_C+R_E & =1.56 \\ R_C & =1.56-1=0.56 \mathrm{k} \Omega \end{aligned} \end{aligned}$$