$z=\frac{{(1+i)}^2}{a-i}\times \frac{a+i}{a+i}$

$z=(\frac{(1-1+2i)(a+i)}{a^2+1}=\frac{2ai-2}{a^2+1}$                       ...(i)

$|z|=\sqrt{{\left(\frac{-2}{a^2+1}\right)}^2+{\left(\frac{2a}{a^2+1}\right)}^2}=\sqrt{\frac{4+4a^2}{{(a^2+1)}^2}}$

$=\sqrt{\frac{4(1+a^2)}{{(1+a^2)}^2}=\frac{2}{\sqrt{1+a^2}}}$

$given|z|=\frac{2}{6}$

$so\sqrt{\frac{2}{5}=}\frac{2}{\sqrt{1+a^2}}$

from equation (i) (square both side)

$\Rightarrow \frac{2}{5}=\frac{4}{1+a^2}$

$\Rightarrow 1+a^2=10$

$a^2=9$

$\Rightarrow a\pm 3\because (a>0)\therefore a=3$

$Hencez=\frac{1+i^2+2i}{3-i}=\frac{2i}{3-i}=\frac{2i(3+i)}{10}=\frac{-1+3i}{5}$

$\overline{z}=\frac{-1}{5}-\frac{3}{5}i$