Thanks Carcass!

hi- before calculating exact values of Quantity A and Quantity B, it is suggested that both the quantities are reduced by common numbers. it will make the calculation easier

Quantity A

As the first step rewrite \(45\) as \(9 \times 5\)

\(\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}\) = \(\frac{75}{4^2} \times \frac{3^2}{9 \times 5} \times \frac{2^4}{ 9 \times 5}\)

and then proceed to cancel all like terms on both sides

\(\frac{75}{4^2} \times \frac{3^2}{3^2 \times 5} \times \frac{2^2 \times 2^2}{ 3^2 \times 5}\) = \(\frac{75 \times 2^2 \times 2^2}{3^4 \times 5^2}
\) = \(\frac{75 \times 2^2}{3^4}\) = \(\frac{75 \times 2^2}{3^3}\) = \(\frac{300}{27}\)

Quantity B

\(\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}\) = \(\frac{2^2}{5^2} \times \frac{10}{3}\) = \(\frac{10}{3}\) = \(10\) = \(10\)

Quantity A: \(\frac{300}{27}\)
Quantity B: \(10\)

Clearly \(\frac{300}{27} > 10\)

The answer is A.