The output ac voltage is \(2.0\,V\). So, the ac collector current \(i _{ C }=2.0 / 2000=1.0\,mA\).

The signal current through the base is, therefore given by

\(i _{ B }= i _{ C } / \beta=1.0\,mA / 100=0.010\,mA\)

The \(dc\) base current has to be \(10 \times 0.010=0.10\) \(mA\).

\(R _{ B }=\left( V _{ BB }- V _{ BE }\right) / I _{ B }\)

Assuming \(V _{ BE }=0.6 V , R _{ B }=(2.0 .0 .6) / 0.10\) \(=14\,k \Omega\)