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(D) The work done $W$ by the electric field on the charge $q$ as it moves from $x = 0$ to $x = x_{0}$ is given by the integral of the force $F = qE$ w

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$\sqrt{\frac{3}{a}}$

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(D) The work done $W$ by the electric field on the charge $q$ as it moves from $x = 0$ to $x = x_{0}$ is given by the integral of the force $F = qE$ with respect to displacement $dx$.$W = \int_{0}^{x_{0}} qE \, dx = qE_{0} \int_{0}^{x_{0}} (1 - ax^{2}) \, dx$Integrating the expression,we get:$W = qE_{0} \left[ x - \frac{ax^{3}}{3} ight]_{0}^{x_{0}} = qE_{0} \left( x_{0} - \frac{ax_{0}^{3}}{3} ight)$According to the work-energy theorem,the change in kinetic energy $\Delta KE$ is equal to the work done $W$. Since the particle starts from rest and we want to find the position where the kinetic energy is zero again,$\Delta KE = 0$,which implies $W = 0$.Setting $W = 0$:$qE_{0} \left( x_{0} - \frac{ax_{0}^{3}}{3} ight) = 0$Since $q 
eq 0$ and $E_{0} 
eq 0$,we have:$x_{0} - \frac{ax_{0}^{3}}{3} = 0$$x_{0} (1 - \frac{ax_{0}^{2}}{3}) = 0$Ignoring the initial position $x_{0} = 0$,we solve for $x_{0}$:$1 - \frac{ax_{0}^{2}}{3} = 0$$ax_{0}^{2} = 3$$x_{0} = \sqrt{\frac{3}{a}}$

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