The charge per unit length of the four quadrant of the ring is 2λ,-2λ,λ and -λ respectively. the electric field at the centre is

I will use ¶ for lambda, and @ for theta,For 2¶, if you consider a linear element (length wise) then it`ll have, say, dx length which would subtend an angle say, d@ at the center and R is the radius. So charge of the element is dq=2¶.dx=2¶R.d@. If you work it out, you shall see that all the sine components along Y axis get cancelled, if you consider an angle @ with the X-axis. So, net electric field along X axis will be integration over dE×cos @ from 0 to π/2. dE= k×dq/R², where k=1/4π€°. On calculating for -2¶,-¶ and ¶ and adding all of them you would get -¶/2π€°R i as answer.