A man is known to speak the truth 3 out of 4 times

Explanation:

Let R,S be the event that the man reports that is a six and the throw is a six respectively.  

P(R|S)= the probability that the man reports it is a six when it is a six = probability of speak truth =34

P(R|S')= the probability that the man reports it is a six when it is not a six = probability of not speak truth =14

P(S)= the probability that throw is six =16

P(S')= the probability that throw is not a six =1-16=6-16=56

Now we have to find the probability that the throw is actually six when the man reports it is a six i.e. P(S|R)

By the Bayes theorem, 

P(S|R)=P(S)P(R|S)P(S)P(R|S)+P(S')P(R|S')

=16341634+5614

=324324+524=3245+324=324248=38

Final Answer:

Hence the correct option is (A).i.e 38