1. Given data

• Infinite number of charges are placed on the positive $y$-axis at $y=1,\,2,\,4,\,8,\,\dots \text{ m}$
• Each charge is $q=1\,\mu C=10^{-6}\,C$
• A test charge of $Q=1\,C$ is placed at the origin.
• Coulomb constant: $k=\frac{1}{4\pi\varepsilon_0}=9\times 10^9\,\text{N m}^2/\text{C}^2$

2. Force due to one charge at distance $r$

The magnitude of force on the $1\,C$ charge due to one source charge is $F=\frac{kqQ}{r^2}$ Since all source charges are positive and lie above the origin, each force on the charge at the origin is along the negative $y$-direction. We only need the total magnitude.

Thus, $F_n=\frac{9\times 10^9\times 10^{-6}\times 1}{r_n^2}$ $F_n=\frac{9\times 10^3}{r_n^2}\,\text{N}$

3. Distances of the charges

The distances are $r_n=1,2,4,8,\dots$ So the total force magnitude is $F=9\times 10^3\left(1+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}+\cdots\right)$

That is, $F=9\times 10^3\left(1+\frac14+\frac1{16}+\frac1{64}+\cdots\right)$

4. Sum the infinite geometric series

This is a geometric progression with $a=1,\qquad r=\frac14$ Hence, $S=\frac{a}{1-r}=\frac{1}{1-\frac14}=\frac{1}{\frac34}=\frac43$

Therefore, $F=9\times 10^3\times \frac43$ $F=12\times 10^3\,\text{N}$

5. Find $x$

Given that total force is $x\times 10^3\,\text{N}$, $x=12$

So, to the nearest integer, $\boxed{12}$

6. Comparison with stored answer

Stored correct answer = $12$.

Our derived answer matches the stored answer.