Radius of the loop, $r=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Total charge on the loop, $Q=1 \times 10^{-6} \mathrm{C}$
$$
\begin{aligned}
& \therefore \text { Charge per unit length, } \lambda=\frac{Q}{2 \pi r}=\frac{10^{-6}}{2 \pi \times 0.01} \\
& =\frac{10^{-4}}{2 \pi} \mathrm{C} / \mathrm{m} \\
& l^{\prime}=0.01 \% \text { of the length }=\frac{0.01}{100} \times 2 \pi r \\
& \Rightarrow \quad l^{\prime}=\frac{2 \pi \times 0.01 \times 0.01}{100} \\
& =2 \pi \times 10^{-6} \mathrm{~m}
\end{aligned}
$$
For the whole circular loop, the electric field at the centre is zero due to symmetry. But when a certain portion of the loop is cut off, then there will exist an electric field at the centre due to that cutting part.
Which is equal to the electric field due to remaining part of wire as per charge conservation.
$\therefore$ Charge of the part $l^{\prime}, Q^{\prime}=l^{\prime} \lambda=\frac{2 \pi \times 10^{-6} \times 10^{-4}}{2 \pi}$
$$
=10^{-10} \mathrm{C}
$$
$\therefore$ Electric field at centre,
$$
\begin{aligned}
E & =\frac{Q^{\prime}}{4 \pi \varepsilon_0 r^2} \\
E & =\frac{10^{-10} \times 9 \times 10^9}{(0.01)^2} \\
& =9 \times 10^3 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$