The correct answer is $\boxed{\frac{5}{{18}}}$.
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
In this case, event A is the student getting above 90% marks, and event B is the student passing the exam. We are given that $P(A \cap B) = \frac{5}{{100}}$ and $P(B) = \frac{20}{{100}}$. Substituting these values into the formula, we get:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{{100}}}{\frac{20}{{100}}} = \frac{5}{{18}}$$
Therefore, the probability that the student gets above 90% marks, given that the student passes the examination, is $\frac{5}{{18}}$.
Option A is incorrect because it is the probability of the student getting above 90% marks, not the conditional probability of the student getting above 90% marks given that the student passes the examination.
Option B is incorrect because it is the probability of the student passing the examination, not the conditional probability of the student getting above 90% marks given that the student passes the examination.
Option C is incorrect because it is the probability of the student getting above 90% marks and passing the examination, not the conditional probability of the student getting above 90% marks given that the student passes the examination.