first we will find the expression for E at the centre due to an arc

resultant E due to the small charges element would be in the direction  . joining the midpoint of the arc to the centre

conider only the half arc , & take only the component parallel to the above mentioned directions

  E =∫ 2/4¶ε dq cos( 45-x)/ r^2                     dq = λdl = λr dx

 E =∫ 2/4¶ε λ cos( 45-x)/ r  dx

 E = λ/2¶ε r ∫cos( 45-x)  dx

 E =  λ/2¶ε r sin(45-x) ]

E =  λ/2¶ε r  *1/√2

charge density at the first arc ,  λ1 = +2Q/ ¶R

E1 = + 2Q/ ¶R 2¶ε R  *1/√2

  = +Q /√2 ε ¶^2 R^2 

λ2 = -4Q / ¶R

E2 = -Q /√2 ε ¶^2 R^2 

λ3 = 6Q/ ¶R

E3 = Q /√2 ε ¶^2 R^2

Resultant E = E1 +E2 +E3

                 =  Q /√2 ε ¶^2 R^2   to the direction joining the midpoint of the arc to the centre & towards the centre