Answer & Solution
Answer: Option D
Solution:
$$\eqalign{
& {I_{\text{C}}} = \frac{{{I_{{\text{CO}}}}}}{{1 - \alpha }} + \frac{\alpha }{{1 - \alpha }} \times {I_\beta } \cr
& = \frac{6}{{1 - 0.98}} + \frac{{0.98}}{{1 - 0.98}} \times 100 \cr
& = 5.2\,{\text{mA}} \cr} $$