Let z be a complex number such that

Let $z=x+i y$

Then, $\left|\frac{z-i}{z+2 i}\right|=1 \Rightarrow x^{2}+(y-1)^{2}$

$=x^{2}+(y+2)^{2} \Rightarrow-2 y+1=4 y+4$

$\Rightarrow \quad 6 y=-3 \Rightarrow y=-\frac{1}{2}$

$\because \quad|z|=\frac{5}{2} \Rightarrow x^{2}+y^{2}=\frac{25}{4}$

$\Rightarrow x^{2}=\frac{24}{4}=6$

$\therefore \quad z=x+i y \quad \Rightarrow \quad z=\pm \sqrt{6}-\frac{i}{2}$

$|z+3 i|=\sqrt{6+\frac{25}{4}}=\sqrt{\frac{49}{4}}$

$\Rightarrow|z+3 i|=\frac{7}{2}$