Answer:

To find $I_B$

Applying KVL to the base -emitter loop , we get

$5=V_{BE}+I_ER_E$

Substitute $I_E=(1+\beta)I_B$

 $\therefore 5-0.7=(1+\beta)I_BR_E$

$\therefore 5-0.7=(1+100)I_B\times 1 \times 10^3$

$\therefore I_B=0.0425mA$

The emitter current can also be found by using ,

$I_E=101 \times 0.0425 \times 10^{-3}=4.29mA$

Now ,  $ I_C=\beta I_B$

$\therefore I_C=100 \times 0.0425 \times 10^{-3}=4.25mA$

Applying KVL to collector loop we get , 

$10-I_CR_C-V_{CE}-I_ER_E=0$

$10=I_CR_C+V_{CE}+I_ER_E$

$10=4.25 \times 0.5+V_{CE}+4.29$

$\therefore V_{CE}=10-2.125-4.29=3.585V$