The order of the prime numbers start from $\displaystyle 2, 3, 5, 7, 11, 13$. $\displaystyle 1$ is not prime as it's a unit. $\displaystyle 4$ is a composite number. So our fourth smallest prime number is $\displaystyle 7$.

Since all the numbers are odd and don't end with a $\displaystyle 5$, let's check the basic divisbility rule. The divisibility rule for $\displaystyle 3$ is if the digits have a sum divisible by $\displaystyle 3$, then it is. 

$\displaystyle 39=3+9=12$

$\displaystyle 31=3+1=4$

$\displaystyle 37=3+7=10$

$\displaystyle 43=4+3=7$

$\displaystyle 53=5+3=8$

Based on this analysis, only $\displaystyle 39$ is divisible by $\displaystyle 3$ and therefore not prime and is our answer. 

Prime numbers are integers. So doing division and square roots will not generate integers. By doing multiplication and exponents, we involve more factors. The only possibility is subtraction. If $\displaystyle p$ was $\displaystyle 7$ and we subtracted $\displaystyle 2$ we get $\displaystyle 5$ which is also prime. 

Other than the number itself and one, we also need to have prime factors in that number. Since it's not a perfect square, we need to find the smallest possible prime numbers. That will be $\displaystyle 2*3$ or $\displaystyle 6$ which is our answer. 

Since all the numbers are odd and don't end with a $\displaystyle 5$, let's check the basic divisbility rule. The divisibility rule for $\displaystyle 3$ is if the digits have a sum divisible by $\displaystyle 3$, then it is. 

$\displaystyle 687=6+8+7=21$

$\displaystyle 787=7+8+7=22$

$\displaystyle 853=8+5+3=16$

$\displaystyle 983=9+8+3=20$

$\displaystyle 691=6+9+1=16$

Based on this analysis, only $\displaystyle 687$ is divisible by $\displaystyle 3$ and therefore not prime and is our answer. 

Let's work backwards. All even numbers are not prime so we skip $\displaystyle 98$. $\displaystyle 99$ is clearly divisible by $\displaystyle 9$. $\displaystyle 95$ is definitely a composite number as it's divisible by $\displaystyle 5$. $\displaystyle 93$ is divisible by $\displaystyle 3$ because of the divisibility rule $\displaystyle (93=9+3=12)$.$\displaystyle 91$ is divisbile by $\displaystyle 7$. The divisibility rule for $\displaystyle 7$  is double the last digit and subtract from the rest $\displaystyle (91=9-1*2=7)$. From the remaining answers, $\displaystyle 97$ is prime and is the largest prime number under $\displaystyle 100$ and is our answer. 

This will require us to know the divisibility rule of $\displaystyle 11$. The reason for this choice is that some of the numbers are palindromes like $\displaystyle 11$ so we eliminate $\displaystyle 77, 99$. For the three digit numbers, the divisibility rule for $\displaystyle 11$ is add the outside digits and if the sum matches the sum then it is divisible. Let's see.

$\displaystyle 101=1+1\neq0$

$\displaystyle 121=1+1=2$

$\displaystyle 132=1+2=3$

Based on this test, $\displaystyle 101$ is not divisible by $\displaystyle 11$ and is our answer.

Plug in a prime number such as $\displaystyle 7$ and evaluate all the possible solutions. Note that the question asks which value COULD be prime, not which MUST BE prime. As soon as your number-picking yields a prime number, you have satisfied the "could be prime" standard and know that you have a correct answer.