Assume the population of city R was 96000 in 2016. Then, in percentage terms, by how much was it more than the male population of city T?

It is given that the population of R in 2016 was 96000, and R contains 25% of the total population.

Let the total population be x.

$$\ \dfrac{\ 25x}{100}=96000$$

$$x=384000$$

T contains 12% of the total population.

Population of T = $$\ \dfrac{\ 12}{100}\times\ 384000\ =46080$$

Male population of T = $$\ \dfrac{\ 60}{100}\times\ 46080\ =27648$$

Percentage difference of population of R from the male population of city T = $$\ \dfrac{\ 96000-27648}{27648}\ =\ \dfrac{\ 68352}{27648}$$
= 247.2%