Here, $\left|z-1-2i\right|=1$ represents a circle with centre $\left(1,2\right)$ and radius $1$ unit.

The complex number $z=\mathrm{x}+i\mathrm{y}$ satisfying the given inequality and having the least positive argument is the point of contact of the tangent from the origin to the circle with the least positive slope.

From the diagram,

$\tan \varphi =\frac{1}{2}$
$\therefore \tan 2\varphi =\frac{2\tan \varphi }{1-{\tan }^2\varphi }=\frac{4}{3}$

For the least positive argument, $\arg \left(z\right)=\theta$ (let)
$\tan \theta =\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{\pi }{2}-2\varphi \right)=\cot 2\mathrm{\varphi }=\frac{3}{4}$

Also, from the diagram,

$x^2+y^2=4$ and $\tan \theta =\frac{y}{x}=\frac{3}{4}$

i.e. $y=\frac{3}{4}x$

$\Rightarrow x^2+\frac{9x^2}{16}=4\Rightarrow x=\frac{8}{5}$

Hence, for the least positive argument, the real part of $z$ is equal to $\frac{8}{5}$