Answer: Applying KVL to the base loop we get , 

$V_{BB}-I_BR_B-V_{BE}=0$

Substituting the given values , 

$4-220I_B-0.7=0$

$\therefore I_B=\dfrac{3.3}{220}$

$\therefore I_B=0.015mA$

Now the collector current can easily calculated by the following formula

$I_C=\beta I_B$

$\therefore I_C=200\times 0.015$

$\therefore I_C=3mA$

Applying KVL to the collector loop , we get

$V_{CC}-I_CR_C-V_{CE}=0$

$10-3 \times 2 =V_{CE}$

$\therefore V_{CE}=4V$