Answer:

As $ {{V}_{BE}}=0 $, the potential drop across $ {{R}_{B}} $ is 10 Volts.

$ {{I}_{B}}=\frac{10}{400\times {{10}^{3}}}=25\mu A $

Now, as $ {{V}_{CE}}=0 $, The potenial drop across $ {{I}_{C}}{{R}_{C}} $ is 10 Volts.

$ {{I}_{C}}=\frac{10}{3\times {{10}^{3}}}=3.33mA $

$ \therefore \beta =\frac{{{I}_{c}}}{{{I}_{B}}}=\frac{333\times {{10}^{-3}}}{25\times {{10}^{-6}}} $

$ \beta =133 $