a
$B _2=\frac{\mu_0 I }{2 \pi x _1}+\frac{\mu_0 I }{2 \pi\left( x - x _1\right)}$ (opposite )
$B_1=\frac{\mu_0 I }{2 \pi x _1}-\frac{\mu_0 I }{2 \pi\left( x - x _1\right)} \text { (same) }$
Case -$1$ When current is in the same direction
$B=B_1=\frac{3 \mu_0 I }{2 \pi x_0}-\frac{3 \mu_0 I }{4 \pi x _0}=\frac{3 \mu_0 I }{4 \pi x _0} $
$R _1=\frac{ mv }{ qB }$
Case-$2$ When current is in oposite direction
$B = B _2=\frac{9 \mu_0 I }{4 \pi x _0} $
$R _2=\frac{ mv }{ qB } $
$\frac{ R _1}{ R _2}=\frac{ B _2}{ B _1}=\frac{9}{3}=3$
