A circular wire-loop of radius a carries a total charge Q uniformly...

The electric field at the centre of a complete uniformly charged circular loop is zero.

By the principle of superposition, the electric field due to the remaining wire is equal in magnitude and opposite in direction to the electric field due to the small cut-off segment.

The linear charge density of the wire is $\lambda = \dfrac{Q}{2\pi a}$.

The charge on the small segment of length $dL$ is $dq = \lambda dL = \dfrac{QdL}{2\pi a}$.

Treating this small segment as a point charge, the magnitude of the electric field at the centre due to it is $E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{dq}{a^2}$.

Substituting the value of $dq$, we get $E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{\dfrac{QdL}{2\pi a}}{a^2} = \dfrac{QdL}{8\pi^2 \varepsilon_0 a^3}$.

Thus, the electric field at the centre due to the remaining wire is $\dfrac{QdL}{8\pi^2 \varepsilon_0 a^3}$.