Given, drag force $F_d$ depends on air density $\sigma$, velocity of ball $v$ and area of cross-section of ball $A$.

$\therefore$$F_d\propto {\sigma }^a{v}^b{A}^c$

Substituting dimensions of different physical quantities, we have

$\left[ML{T}^{-2}\right]={\left[M{L}^{-3}\right]}^a{\left[L{T}^{-1}\right]}^b{\left[ L^2\right]}^c$

Equating dimensions of fundamental quantities, we have

$a=1....\left(\mathrm{i}\right)$

$-3a+b+2c=1....\left(\mathrm{ii}\right)$

$-b=-2\Rightarrow b=2....\left(\mathrm{iii}\right)$

Putting the values of $a$ and $b$ in Eq (ii), we get

$\Rightarrow -3\times 1+2+2c=1\Rightarrow c=1$

$\because a=1,b=2,c=1$

So, drag force on football is

$F_d=k·\sigma ·v^2·A$

When football reaches terminal speed, its weight is balanced by drag force.

$\Rightarrow$$mg=F_d$

$\Rightarrow$$mg=k\sigma v_T^{2}A$

where, $u_T=$terminal speed of football.

$\Rightarrow mg=k\sigma v_T^2\pi R^2$

$\Rightarrow mg=k\sigma u_T^2\pi {\left(\frac{m}{\frac{4}{3}\pi \rho }\right)}^{\frac{2}{3}}$

[ from $m=\frac{4}{3}\pi R^3\rho$ where $\rho =$ density of football]

$\Rightarrow v_T\propto m^{1/6}$

$\Rightarrow \frac{v_1}{v_2}={\left(\frac{m_1}{m_2}\right)}^{\frac{1}{6}}={\left(\frac{250}{125}\right)}^{\frac{1}{6}}=2^{1/6}$