(A) At a fixed temperature,the average kinetic energy $\frac{1}{2} m \langle v^2 \rangle$ is constant. Thus,the average speed $v_{avg} \propto \frac{1}{\sqrt{m}}$.
The molar mass of $UF_6$ with $U^{235}$ is $M_1 = 235 + 6 \times 19 = 235 + 114 = 349 \text{ units}$.
The molar mass of $UF_6$ with $U^{238}$ is $M_2 = 238 + 6 \times 19 = 238 + 114 = 352 \text{ units}$.
Since $M_1 < M_2$,the $UF_6$ gas containing $U^{235}$ will have the larger average speed.
The ratio of speeds is given by $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{352}{349}} \approx \sqrt{1.008596} \approx 1.004288$.
The percentage difference is $\frac{\Delta v}{v} \times 100 = (1.004288 - 1) \times 100 \approx 0.43 \%$.