Let define the following events:
$\mathrm{E}_1$ : Students gusses the answer to the question.
$\mathrm{E}_2$ : Students copies the answer to the question.
$\mathrm{E}_3$ : Students know the answer to the quesiton.
$\mathrm{E}:$ Answer is correct.
Here, $\mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{6}$
$\therefore \mathrm{P}\left(\mathrm{E}_3\right)=1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}$
Since, the question is a multiple choice question with four choice. So the probability that the answer is correct when it is gussed, $\Rightarrow P\left(\frac{E}{E_1}\right)=\frac{1}{4}$
Also, the probability that his answer is correct, given that he copied it is $\frac{1}{8} \Rightarrow P\left(\frac{E}{E_2}\right)=\frac{1}{8}$
Moreover, his answer is correct, given he know the answer is sure event, so $\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right)=1$
Hence, by Baye's theorem, the required probability is
$\Rightarrow P\left(\frac{E_3}{E}\right)=\frac{P\left(\frac{E}{E_3}\right) P\left(E_3\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)+P\left(\frac{E}{E_3}\right) P\left(E_3\right)}$
$\Rightarrow P\left(\frac{E_3}{E}\right)=\frac{1 \times \frac{1}{2}}{\frac{1}{4} \times \frac{1}{3}+\frac{1}{8} \times \frac{1}{6}+1 \times \frac{1}{2}}=\frac{24}{29}$