Let

• $K$ = event that the student knows the answer,
• $G$ = event that the student guesses the answer,
• $C$ = event that the student's answer is correct.

Since the student either knows or guesses, $P(K)+P(G)=1.$

We are given:

1. $P(C\mid G)=\frac{1}{2}$
2. $P(G\mid C)=\frac{1}{6}$
3. If the student knows the answer, he is always correct, so $P(C\mid K)=1.$

We need to find $P(K)$.

Step 1: Let $P(K)=p$

Then $P(G)=1-p.$

Now compute $P(C)$ using total probability: $P(C)=P(C\mid K)P(K)+P(C\mid G)P(G).$ Substituting the given values, $P(C)=1\cdot p+\frac{1}{2}(1-p).$ So, $P(C)=p+\frac{1}{2}-\frac{p}{2}=\frac{1+p}{2}.$

Step 2: Use the condition $P(G\mid C)=\frac{1}{6}$

By conditional probability, $P(G\mid C)=\frac{P(G\cap C)}{P(C)}.$

Now, $P(G\cap C)=P(C\mid G)P(G)=\frac{1}{2}(1-p).$ Hence, $\frac{\frac{1}{2}(1-p)}{\frac{1+p}{2}}=\frac{1}{6}.$

Cancel $\frac{1}{2}$ from numerator and denominator: $\frac{1-p}{1+p}=\frac{1}{6}.$

Step 3: Solve for $p$

Cross-multiplying, $6(1-p)=1+p.$ So, $6-6p=1+p,$ $5=7p,$ $p=\frac{5}{7}.$

Thus, $P(K)=\frac{5}{7}.$

Step 4: Match with options

The correct option is: $\boxed{\text{C } \frac{5}{7}}$