If the probability that the product of the outcomes of three  rolls of a fair dice is a prime number p then the value of  3p is

$\text{ Total number of cases }n\left(S\right)=6^3=216$

Product is prime only when two outcomes are 1 and the third is prime i.e. 2,3,,5. 
If it is 2, 1, 1 then number of cases are 3 

Similarly 3 cases, for 3, 1, 1 and 5, 1, 1 each

 Hence favorable cases = 9 

$\text{ Hence required probability }p=\frac{1}{24}$