Given that,
$$
\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2
$$
Since, $\log _a b>c, b < a^c$ if $0 < a < 1$
$$
\begin{aligned}
& \Rightarrow \quad \frac{|z|^2-|z|+1}{2+|z|} < \left(\frac{1}{\sqrt{3}}\right)^{-2} \\
& \Rightarrow \frac{|z|^2-|z|+1}{2+|z|} < (\sqrt{3})^2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad|z|^2-|z|+1 < (\sqrt{3})^2(2+|z|) \\
& \Rightarrow \quad|z|^2-|z|+1 < 6+3|z| \\
& \Rightarrow \quad|z|^2-4|z|+1 < 6 \\
& \Rightarrow \quad(|z|-2)^2 < 9 \Rightarrow-3 < |z|-2 < 3 \\
& \Rightarrow \quad-1 < |z| < 5 \Rightarrow 0 < |z| < 5
\end{aligned}
$$
Hence, $\mathrm{z}$ lies inside the circle.