Please solve this magnetism question?

Apply right handed cork screw rule to determine the direction of the `vec B`

So,at point `A` i.e midway between the the two wire,magnetic field due to wire `1` is `-(2mu_0 i)/(4 piR)` direction is going into the plane of the page.

and for wire `2` it is `(2mu_0 i)/(4 piR)` as well but direction is coming out of the page of the plane,

So,magnetic field at point `A` is `(2mu_0 i)/(4 piR)-(2mu_0 i)/(4 piR)=0`

At point `B` magnetic field due to wire `1` is `B_1=(2mu_0 i)/(4 piR)` coming out of the plane of the page, and due to wire `2` it is `B_2=(2mu_0 i)/(4 pi*3R)= (mu_0 i)/(6 piR)` (as here distance is `R+R/2=(3R)/2`) direction is coming out of the plane of page.

So,net magnetic field at point `B` is `B_1 +B_2=B` (let)

Again,for point `C` due to wire `2` magnetic field `B_2' =-(2mu_0 i)/(4 piR)` direction is going into the plane of the page,and due to wire `1` it is `B_1'= -(mu_0 i)/(6 piR)` (compare for wire `2` effect on point `A`) and the direction is going into the plane of the page.

So,net magnetic field at point `C` is `B_1'+B_2'=B'`(let)

Clearly, `B=-B'` (direction is opposite of each other)

So,the ratio of magnetic field at point `A` ,`B` and `C` is `0:1:-1`