Given $\left|\frac{z-i}{z+2i}\right|=1$ This implies $|z-i|=|z+2i|$ because the modulus of a quotient is the quotient of moduli.

Let $z=x+iy$ Then $z-i=x+i(y-1), \qquad z+2i=x+i(y+2)$ So, $|z-i|^2=x^2+(y-1)^2$ $|z+2i|^2=x^2+(y+2)^2$

Since these are equal, $x^2+(y-1)^2=x^2+(y+2)^2$ $(y-1)^2=(y+2)^2$ $y^2-2y+1=y^2+4y+4$ $-6y=3$ $y=-\frac12$

Also given $|z|=\frac52$ Hence, $x^2+y^2=\left(\frac52\right)^2=\frac{25}{4}$ Substituting $y=-\frac12$, $x^2+\frac14=\frac{25}{4}$ $x^2=6$

Now find $|z+3i|$. Since $z+3i=x+i\left(y+3\right)=x+i\left(-\frac12+3\right)=x+i\frac52$ therefore, $|z+3i|^2=x^2+\left(\frac52\right)^2$ $=6+\frac{25}{4}=\frac{49}{4}$ So, $|z+3i|=\frac72$

Comparing with options: $\frac72$ corresponds to Option D.