Correct Answer - B
`log_(3)C=3+log_(3)a`
`rArr "log"_(3)(c )/(a)=3`
`rArr c = 27 a` ….(1)
`log_(a)b=2` and `log_(b)c=2`
`rArr log_(a)b.log_(b)c=4`
`rArr log_(a)c=4`
`rArr c = a^(4)` ….(2)
From (1) and (2), we get
a = 3 and c = 81
`therefore` from `log_(a)b = 2`, we get `b = a^(2) = 9`
`rArr (c )/(ab)=3`