(R∩S)∪(S∩T)∪(R∩S'∩T') 

R∩S is made up of the two regions which are 
in common to the circles R,S.  They are
regions 2&5, So we can substitute 2&5 for R∩S.

(2&5)∪(S∩T)∪(R∩S'∩T')

S∩T is made up of the two regions which are 
in common to the circles S,T.  They are
regions 5&6, So we can substitute 5&6 for S∩T.

(2&5)∪(5&6)∪(R∩S'∩T')

R∩S'∩T' is a little more complicated. There are 
primes (tic marks) on S and T, indicating
complements.  That means that R∩S'∩T' is the 
part of R that is NOT part of S, and also NOT 
part of T.  That is only the left part of circle 
R, the one region 1, because it is the only part 
of R that is NOT part of the other two circles.  
So we replace R∩S'∩T' by 1.
  
(2&5)∪(5&6)∪(1)

So we end up with the 4 regions 1&2&5&6, so

(R∩S)∪(S∩T)∪(R∩S'∩T') = 1&2&5&6

Regions 1 and 2 are the parts of R that are not parts
of T. So that's (R∩T')

We only need to union that with regions 5 and 6.
Regions 5 and 6 are the regions common to S and T

(R∩T')∪(S∩T)

Edwin