The Base Voltage of a BJT transistor in Voltage divider bias configuration is given by

$V_{b}=V_{th}=V_{cc}( \dfrac{R2}{R1+R2})$

$V_{b}=20 \dfrac {2.2}{8.2+2.2}$ 

     =$20\dfrac{2.2}{10.4} $

     =4.320V

The value of Threshold Resistor is parallel combination of R1 and R2.that is

$R_{th}=\dfrac{R1R2}{(R1+R2)}$

       =$\dfrac{8.2\times2.2}{8.2+2.2}$

       =1.734$\Omega$

$I_{c}=\beta{I_{b}}$

   =$\dfrac{\beta(V_{th}-V_{be})}{R_{th}+(\beta+1)R_{E})}$

   =$120\dfrac{(4.32-(4.32+20)}{(1.734+(121\times1800)}$

   =$120\dfrac{(4.32-24.32)}{(1.734+217800)}$

   =$\dfrac{2400}{217801.734}$

   =11mA