The number of those who eat breakfast, or floss regularly, or get the morning paper 
(who do any (~ at least one) of these three morning activity) is


    115 - 14 = 101.

Let B is the set of those who eat breakfast   (and the number of people in this set, at the same time, which is 85 - given !).

Let F is the set of those who floss regularly (and the number of people in this set, at the same time, which is 58 - given !).

Let P is the set of those who subscribe paper (and the number of people in this set, at the same time, which is 27 - given !).


Let BF is the intersection of the sets B and F (and the number of people in this intersection, at the same time, which is 52 - given !).

Let BP is the intersection of the sets B and P (and the number of people in this intersection, at the same time, which is 15 - given !).


Let BFP is the intersection of the sets B, F and P (and the number of people in this intersection, at the same time, which is 10 - given !).



The number of those who do any of these morning activities is 115 - 14 = 101.



This number, which is the number of people in the union of the sets B, F and P, can be calculated in other way, too, using the formula


   101 = n(B U F U P) = B + F + P - BF - BP - FP + BFP.    *()


In this equation, we know every term of the right side, except FP (which is under the question b) ).

So, we substitute all known values into equation (*), and we get


    101 = 85 + 58 + 27 - 52 - 15 - FP + 10.


Now we can easily find the value of FP as


    FP = 85 + 58 + 27 -52 - 15 + 10 - 101 = 12.


Answer.  The number of those who use dental floss regularly and also get the morning paper is 12.