NOTE:
$Source: Ullman $
Easy method:
As you can see $First(A) = \{a, c, \varepsilon\} = \{First(B), \varepsilon\}$
from this we can say that other production of A should be starting with terminal B
and there is only one option matching this case $Option C$
$A\rightarrow BbS | \varepsilon$
gives $First(A) = \{First(BbS), \varepsilon\} = \{First(B), \varepsilon\} = \{a, c, \varepsilon\}$
so $Answer: C$
Using Option Verification:
