We know, radius of the orbit is given by :
[tex]r=\dfrac{mv}{qB}[/tex]
So, ratio is given by :
[tex]\dfrac{q}{m}=\dfrac{v}{Br}\\\\\dfrac{q}{m}=\dfrac{109\ m/s}{0.0691 \ T \times 0.0427\ m}\\\\\dfrac{q}{m}=36942.01 \ C/kg\\\\\dfrac{q}{m}=3.69\times 10^{4}\ C/kg[/tex]
Therefore, the charge–to–mass ratio of this particle is [tex]3.69\times 10^{4}\ C/kg[/tex] .
Hence, this is the required solution.