Step 1: Circuit Diagram

Step 2: Finding Vcc

$Vcc = 2V_{CEQ}$

Vcc = 10V

Step 3: Finding RE

$V_{RE} = 10% of Vcc$

$V_{RE} = 1V$

We know that,

$V_{RE} = I_ER_E$

Finding $I_E,$

$I_C= \beta I_B$

$I_B = I_C/\beta = 5mA/100 ………………..Ic=5mA(Given)$

$I_B= 50\mu A$

$I_E = I_B+I_C$

$I_E = 5mA+50\mu A$

$I_E= 5.05mA$

$V_{RE} = I_ER_E$

$1V = 5.05mA×R_E$

$R_E=198.01\Omega$

Step 4: Finding RC

Apply KVL from Vcc to Gnd through collector emitter,

$Vcc - I_CR_C- V_{CEQ} - V_{RE}=0$

$10V – (5mA×R_C) – 5V – 1V=0$

$R_C=800 \Omega$

Step 5: Finding R1 and R2

$I1 = 10 IB$

$I1 = 500µA$

Apply KCL at base of transistor,

$I2 = I1 - I_B$

$I2 = 500\mu A-50\mu A$

$I2= 450\mu A$

Apply KVL from R2 to ground through base to emitter,

$V_{R2} – V_{BE} – V_{RE} = 0$

$V_{R2} = 0.7V + 1V$ ……………Assume $V_{BE}=0.7V$

$V_{R2}= 1.7V$

Hence,

$V_{R2} = I2 R2$

$1.7V = 450\mu A\times R2$

$R2= 3.77K\Omega$

$R1=\frac{Vcc-V_R2}{I1}$

$R1=\frac{10V-1.7V}{500\mu A}$

$R1=16.6K\Omega$

Step 6: Designed Circuit Diagram