Given, Binomial expansion

$${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$$

$$\therefore$$ General Term

$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$$

$$ = {}^{15}{C_r}\,.\,{2^{15 - r}}\,.\,{x^{{1 \over 5}(15 - r - r)}}\,.\,{( - 1)^r}$$

For $${x^{ - 1}}$$ term;

$${1 \over 5}(15 - 2r) = - 1$$

$$ \Rightarrow 15 - 2r = - 5$$

$$ \Rightarrow 2r = 20$$

$$ \Rightarrow r = 10$$

m is the coefficient of $${x^{ - 1}}$$ term,

$$\therefore$$ $$m = {}^{15}{C_{10}}\,.\,{2^{15 - 10}}\,.\,{( - 1)^{10}}$$

$$ = {}^{15}{C_{10}}\,.\,{2^5}$$

For $${x^{ - 3}}$$ term;

$${1 \over 5}(15 - 2r) = - 3$$

$$ \Rightarrow 15 - 2r = - 15$$

$$ \Rightarrow 2r = 30$$

$$ \Rightarrow r = 15$$

n is the coefficient of $${x^{ - 3}}$$ term,

$$\therefore$$ $$n = {}^{15}{C_{15}}\,.\,{2^{15 - 15}}\,.\,{( - 1)^{15}}$$

$$ = 1\,.\,1\,.\, - 1$$

$$ = - 1$$

Given,

$$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$$

$$ \Rightarrow {}^{15}{C_{10}}\,.\,{2^5}\,.\,{(1)^2} = {}^{15}{C_r}\,.\,{2^r}$$ [putting value of m and n]

$$ \Rightarrow {}^{15}{C_{15 - 10}}\,.\,{2^5} = {}^{15}{C_r}\,.\,{2^r}$$

$$ \Rightarrow {}^{15}{C_5}\,.\,{2^5} = {}^{15}{C_r}.\,{2^r}$$

Comparing both side, we get

$$r = 5$$.