Correct Option: B

Given equation is,
( 6a^-2bc^-3/4ab^-3c² ) ÷ ( 5a^-3 b²c^-1/3ab^-2c³ )
Apply the algebra law a ÷ b = a x 1/b
= ( 6a^-2bc^-3/4ab^-3c² ) x ( 3ab^-2c³/5a^-3 b²c^-1 )
= ( 6a^-2 bc^-3 x 3ab^-2c³) / ( 4ab^-3c² x 5a^-3b²c^-1 )
Apply the law of Fractional Exponents and Laws of Exponents
(a^m)(aⁿ) = a^m+n
a^m÷aⁿ=a^m-n
Or
a^m/aⁿ=a^m-n
= 18a^{-2 + 1} b^{1 - 2} c ^{-3 + 3} / 20a^{1 - 3} b^{-3 + 2} c^{2 - 1}
= 9a^-1b^-1c⁰/ 10a ^{- 2} b^-1c¹
= ( 9/10 ) a^{-1 + 2}b^{-1 + 1} c^{0 - 1}
= ( 9/10 ) a¹b⁰ c^-1
= ( 9/10 ) ac^-1 [∵ b⁰ = 1]