Let $A$ be the event that the examinee gives the correct answer.

Let $G, C, K$ stand for guessing, copying and knowing. respectively.

Given, $P\left(C\right)=\frac{1}{6}, P\left(G\right)=\frac{1}{3}$

and $P\left(\frac{A}{C}\right)=\frac{1}{8}$

$\Rightarrow P\left(C\right)+P\left(G\right)+P\left(K\right)=1$

$\Rightarrow \frac{1}{6}+\frac{1}{3}+P\left(K\right)=1$

$\therefore P\left(K\right)=\frac{1}{2}$

Also, $P\left(\frac{A}{K}\right)=1,$ for if the examinee knows, he/she will correctly answer it and $P\left(\frac{A}{G}\right)=\frac{1}{4},$

 since there are four choices.

Now, total probability $P\left(A\right)=P\left(G\right)P\left(\frac{A}{G}\right)+P\left(C\right)P\left(\frac{A}{C}\right)+P\left(K\right)P\left(\frac{A}{K}\right)$

$=\frac{1}{3}·\frac{1}{4}+\frac{1}{6}·\frac{1}{8}+\frac{1}{2}·1$

$=\frac{1}{12}+\frac{1}{48}+\frac{1}{2}=\frac{4+1+24}{48}=\frac{29}{48}$

Now, $P\left(\frac{K}{A}\right)=\frac{P\left(K\right)P\left(\frac{A}{K}\right)}{P\left(A\right)}$

$=\frac{\frac{1}{2}·1}{\frac{29}{48}}=\frac{1}{2}·\frac{48}{29}=\frac{24}{29}$