Let's divide the loop in two parts. 1. The small element of length, $\mathrm{d}l$, which is to be cut off, 2. The Remaining part of the loop, 
Initially, by symmetry, the net Electric field at the center is zero.
∴ $E$ Remaining Wire ​ $+Edl$ element ​$=0$,
$\Rightarrow E$ remaining wire$=-E\mathrm{d}l$ element .
The charge per unit length of the wire is, $\lambda =\frac{Q}{2\pi a}$, the charge on the $\mathrm{d}l$ length of the wire, $\mathrm{d}Q=\lambda \mathrm{d}l=\frac{Q\mathrm{d}l}{2\pi a}$. 
​​​​​​The small element can be considered as a point charge, So its Electric Field at the center, 
$E_{dl}=\frac{kdQ}{a^2}=\frac{kQdl}{2\pi a^3}=\frac{9\times {10}^9\times {10}^{-5}\times \pi \times {10}^{-6}}{2\times \pi \times {\left(10\times {10}^{-2}\right)}^3}=45 \mathrm{N} {\mathrm{C}}^{-1}$