Two long straight wires P and Q carrying equal current 10A each...

As we know, force per unit length between two parallel straight wires is given by, $F = \dfrac{{\mu}_{0}{i}_{1}{i}_{2}}{2\pid}$.

Therefore, required ratio

$\dfrac{{F}_{1}}{{F}_{2}} = \dfrac{\dfrac{{\mu}_{0} ( 10{)}^{2}}{2\pi ( 5 \times {10}^{- 2} )}}{\dfrac{{\mu}_{0} ( 20{)}^{2}}{2\pi\left(\dfrac{5}{2} \times {10}^{- 2}\right)}} = \dfrac{1}{8}$

$⇒ {F}_{2} = 8{F}_{1}$