Answer
Student 3 heat engine design is the only acceptable one.
Work Step by Step
First of all, we need to find the maximum efficiency of this heat engine which is the Carnot's heat engine's efficiency, which is given by
$$\eta_{\rm Carnot}=1-\dfrac{T_C}{T_H}$$
Plugging the known;
$$\eta_{\rm C}=1-\dfrac{300}{500}=\bf 0.4$$
Now if the efficiency of this heat engine is less than this Carnot's efficiency then it is a suitable one and if it exceeds this value then there is something wrong in the student's data.
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$\bullet$1) Now let's test the efficiency of this heat engine given by the first student:
$$\eta_1=\dfrac{W_{out}}{Q_H}=\dfrac{110}{250}=\bf 0.44$$
which is greater than $\eta_{\rm Carnot}$ and this violates the second law of thermodynamics.
So, it is not acceptable, and hence no further math is needed here.
$\bullet$2) For the second student:
$$\eta_2=\dfrac{W_{out}}{Q_H}=\dfrac{90}{250}=\bf 0.36$$
which is less than $\eta_{\rm Carnot}$ and this suits the second law of thermodynamics. So we have to test the result of the first law.
$$Q_H=W_{out}+Q_C=90+170=\bf 260\;\rm J$$
which is greater than the given $Q_H=\bf 250\;\rm J$ and this violates the first law of thermodynamics.
So, the design is not acceptable.
$\bullet$2) For the third student:
$$\eta_3=\dfrac{W_{out}}{Q_H}=\dfrac{90}{250}=\bf 0.36$$
which is less than $\eta_{\rm Carnot}$ and this suits the second law of thermodynamics. So we have to test the result of the first law.
$$Q_H=W_{out}+Q_C=90+160=\bf 250\;\rm J$$
which is equal to the given $Q_H$ and hence this design is not violating any of the two thermodynamics laws.
Therefore, the third design of student 3 is the only acceptable one.
