1. Let $z=x+iy \quad \Rightarrow \quad \overline z = x-iy$ where $x,y\in\mathbb R$.

2. The given equation is $z^2+3\overline z=0.$ Substitute $z=x+iy$: $(x+iy)^2+3(x-iy)=0.$

3. Expand: $(x^2-y^2+2ixy)+3x-3iy=0.$ So, $(x^2-y^2+3x)+i(2xy-3y)=0.$

4. Equating real and imaginary parts: $x^2-y^2+3x=0$ $2xy-3y=0.$ The second equation gives $y(2x-3)=0.$ Hence, either:

• Case 1: $y=0$
• Case 2: $2x-3=0 \Rightarrow x=\frac32$

5. Case 1: $y=0$

Then the real-part equation becomes $x^2+3x=0$ $x(x+3)=0.$ So, $x=0 \quad \text{or} \quad x=-3.$ Thus solutions are $z=0,\,-3.$

6. Case 2: $x=\frac32$

Then $x^2-y^2+3x=0$ $\left(\frac32\right)^2-y^2+3\left(\frac32\right)=0$ $\frac94-y^2+\frac92=0$ $\frac{27}{4}-y^2=0$ $y^2=\frac{27}{4}$ $y=\pm \frac{3\sqrt3}{2}.$ Thus two more solutions are $z=\frac32 \pm i\frac{3\sqrt3}{2}.$

7. Therefore, total number of solutions is $n=4.$

8. Now evaluate $\sum_{k=0}^{\infty} \frac{1}{n^k}=\sum_{k=0}^{\infty} \left(\frac14\right)^k.$ This is an infinite GP with first term $1$ and common ratio $\frac14$. Hence, $S=\frac{1}{1-\frac14}=\frac{1}{\frac34}=\frac43.$

9. So the required value is $\boxed{\frac43}.$ This corresponds to option B.