$( R _{\text {out }}=200 \Omega, R _{\text {in }}=100 k \Omega,$$ V _{ cC }=3 volt , V _{ BE }=0.7 volt ,V _{ GE }=0, \beta=200 )$
From the above figure,
\(V _{ CE }= V _{ CC }- I _{ C } R _{0}\)
\(0=3- I _{ C }(200)\)
\(I _{ C }=\frac{3}{200}\)
\(=15 \,mA\)
As we know that,
\(\beta=\frac{I_{C}}{ I _{B}}\)
\(I_{B}=\frac{15 mA }{200}\)
\(=75 \mu A\)
Therefore,
\(V _{ BE }= V _{ BB }- I _{ B } R _{ in }\)
\(V _{ BB }=0.7+\left(75 \times 10^{-6}\right)\left(100 \times 10^{3}\right)\)
\(=8.2 V\)