Recycle unreacted ethene
As much as I understand, there is nothing wrong with the question, if your translation is correct. Let's see the reaction:

$$\ce{2 C2H4 (g) + O2 (g) -&gt; 2 C2H4O (g)}$$

The flow reactor was fed with constant flow rates of $\ce{C2H4}$ at $\pu{200 kmol\:h-1}$ and $\ce{O2}$ at $\pu{90 kmol\:h-1}$. Only $6\%$ of the ethene reacts in the reactor. That means the reaction rate of the reactor is $\pu{12 kmol\:h-1}$ per $\ce{C2H4}$. Therefore, assuming a perfect world, $\pu{6 kmol}$ of $\ce{O2}$ would be consumed in each hour to produce $\pu{12 kmol}$ of $\ce{C2H4O}$ as well (see the chemical equation). 

a) Thus, the gas outlet of the reactor would release 98% of unreacted $\ce{C2H4}$ ($0.98 \times \pu{188 kmol}$) and $\ce{O2}$ ($\pu{84 kmol}$), and $\ce{C2H4O}$ ($\pu{12 kmol}$) in each hour as a mixture. The sum ($188 \times 0.98+84+12 \approx \pu{280.2 kmol\:h-1}$) is the rate of gas flow out of filtrator of the flow reactor. Of course, this is assuming filtrator does not filtered off any of these materials (it wasn't clear on the translation). If filtrator separates the product and oxygen from the mixture (e.g., by gas partition by difference in melting points; melting points of oxirane and ethene are ~$\pu{161 K}$ and ~$\pu{104 K}$, respectively, and melting point of oxygen is ~$\pu{54.4 K}$), the outlet rate of $\ce{C2H4}$ out of the filtrator would be $\pu{188 \times 0.98 kmol\:h-1} \approx \pu{184.2 kmol\:h-1}$ (with no recycling of $\ce{C2H4}$). If  the flow of gas out of the filtrator means only the gases pass through the filtrator, that should be only oxygen and oxirane. Thus, the outlet rate of gases out of the filtrator would be $\pu{(84 + 12) kmol\:h-1} = \pu{96 kmol\:h-1}$.

b) I believe, the recycling of $\ce{C2H4}$ means unreacted $\ce{C2H4}$ is adding to original rate of $\pu{200 kmol\:h-1}$. Thus, rate of $\ce{C2H4}$ fed into reactor after first hour is: $\pu{(200+184.2) kmol\:h-1} = \pu{384.2 kmol\:h-1}$. Only 6% of it ($\pu{23.05 kmol\:h-1}$) reacts with $\pu{11.53 kmol\:h-1}$ of oxygen to give $\pu{23.05 kmol\:h-1}$ of oxirane. Therefore, outlet rate is:

$$\left((384.2 - 23.05) + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{462.67 kmol \: h-1}$$

This would be the outlet rate without filtration. If you have to consider only 98% of $\ce{C2H4}$, then the rate would be:

$$\left((384.2 - 23.05)\times 0.98 + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{455.45 kmol \: h-1}.$$