Correct Answer - `2*56xx10^(-20)N`, towards the conductor
`B=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx2)/(0*1)=4xx10^-6T`
When electron travles parallel to the conductor, `vecB` is perpendicular to `vecv`(i.e. `theta=90^@`)
`:. F=evBsintheta`
`=1*6xx10^(-19)xx4xx10^4xx4xx10^-6xxsin90^@`
`=2*56xx10^(-20)N`
This force will be acting towards the conductor.