In an entrance test there are multiple choice questions. There are...

Let $E_1$ be the event that the student knows the answer and $E_2$ be the event that the student guesses the answer.

Let $A$ be the event that the student gets the correct answer.

From the given information:
$P(E_1) = \dfrac{90}{100} = \dfrac{9}{10}$
$P(E_2) = 1 - P(E_1) = \dfrac{1}{10}$

If the student knows the answer, the probability of getting it correct is $1$, so $P(A|E_1) = 1$.

If the student guesses, since there are $4$ options with only one correct, $P(A|E_2) = \dfrac{1}{4}$.

The probability that the student was guessing given that the answer is correct is $P(E_2|A)$.

Using Bayes' theorem:
$P(E_2|A) = \dfrac{P(E_2) P(A|E_2)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$

$P(E_2|A) = \dfrac{\dfrac{1}{10} \times \dfrac{1}{4}}{\dfrac{9}{10} \times 1 + \dfrac{1}{10} \times \dfrac{1}{4}}$

$P(E_2|A) = \dfrac{\dfrac{1}{40}}{\dfrac{9}{10} + \dfrac{1}{40}} = \dfrac{\dfrac{1}{40}}{\dfrac{37}{40}} = \dfrac{1}{37}$

Answer: $\dfrac{1}{37}$