ans of  figure 1

magnetic field due to any arc = u_oI@/4pia   .............1

@ is angle substended at the point , a = radus of arc ..

magnetic field due to straight wire = (u_oI/4pia).[sin@₁+sin@₂]    ...........2

angle @ substended by circular arc = (2pi - 2@)

B_arc (field due to arc) = u_oI(2pi-2@)/4piR

                             = u_oI(pi-@)/2piR                ...................3

B_line(field due to straight wire) = (u_oI/4piRcos@). [ 2sin@ ]           ...................4

perpendicula distance = Rcos@

total magnetic field will be summation of these two ,

B(total) = u_oI/2piR. [ pi-@ + tan@]

this is the required ans

in figure 2

magnetic field due to arc whose radius is a  will be

B_a = u_oI(2pi-@)/4pi(a)   

 magnetic field due to arc whose radius is b will be

B_b = u_oI(@)/4pib

net magnetic field = B = sum of both

         B = u_oI/4pi .  [ (2pi-@)/a + @/b ]     

this is the required ans