1. Define events

Let:

• $A$ = urn $A$ is selected
• $B$ = urn $B$ is selected
• $C$ = urn $C$ is selected
• $K$ = the drawn ball is black

Since one urn is selected at random, $P(A)=P(B)=P(C)=\frac{1}{3}.$

2. Find probability of drawing a black ball from each urn

• Urn $A$ has $7$ red and $5$ black balls, total $12$. $P(K\mid A)=\frac{5}{12}.$

• Urn $B$ has $5$ red and $7$ black balls, total $12$. $P(K\mid B)=\frac{7}{12}.$

• Urn $C$ has $6$ red and $6$ black balls, total $12$. $P(K\mid C)=\frac{6}{12}=\frac{1}{2}.$

3. Find total probability of drawing a black ball

By the law of total probability, $P(K)=P(A)P(K\mid A)+P(B)P(K\mid B)+P(C)P(K\mid C).$

So, $P(K)=\frac{1}{3}\cdot\frac{5}{12}+\frac{1}{3}\cdot\frac{7}{12}+\frac{1}{3}\cdot\frac{1}{2}.$

P(K)=\frac{5}{36}+\frac{7}{36}+\frac{1}{6}=rac{12}{36}+\frac{6}{36}=\frac{18}{36}=\frac{1}{2}.

4. Apply Bayes' theorem

We need $P(A\mid K)$: $P(A\mid K)=\frac{P(A)P(K\mid A)}{P(K)}.$

Substitute values: $P(A\mid K)=\frac{\frac{1}{3}\cdot\frac{5}{12}}{\frac{1}{2}}=\frac{5}{36}\cdot 2=\frac{10}{36}=\frac{5}{18}.$

5. Match with the options

$\boxed{\frac{5}{18}}$ So the correct option is C.