Ten positively charged particles are held fixed along the x -axis...

The position of the $n^{\text{th}}$ particle is $x_n = n \times 10 \text{ cm} = n \times 0.1 \text{ m}$.

The charge of the $n^{\text{th}}$ particle follows the pattern $q_n = n^3 \times 10^{-8} \text{ C}$.

The electric force exerted by the $n^{\text{th}}$ particle on a $1 \text{ C}$ charge at the origin is given by Coulomb's law:

$F_n = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q q_n}{x_n^2}$

Substituting the given values:

$F_n = (9 \times 10^9) \dfrac{1 \times (n^3 \times 10^{-8})}{(n \times 0.1)^2}$

$F_n = (9 \times 10^9) \dfrac{n^3 \times 10^{-8}}{n^2 \times 10^{-2}}$

$F_n = 9 \times 10^9 \times n \times 10^{-6} = 9 \times 10^3 n \text{ N}$

The total electric force $F$ is the sum of the forces from all $10$ particles:

$F = \sum_{n=1}^{10} F_n = \sum_{n=1}^{10} 9 \times 10^3 n$

$F = 9 \times 10^3 \sum_{n=1}^{10} n$

Using the formula for the sum of the first $n$ natural numbers, $\sum_{n=1}^{10} n = \dfrac{10 \times 11}{2} = 55$:

$F = 9 \times 10^3 \times 55$

$F = 495 \times 10^3 \text{ N}$

$F = 4.95 \times 10^5 \text{ N}$