part B M = 1 & N = 15
for first digraph
R_{1 = }{ (a,b),(b,c),(c,a)}
R_{2 }= {(a,c),(b,a),(c,b)}
R_{3 = }{ (a,b),(b,c),(c,a)}
so for first digraph m=1 and n = 3
similarly for second digraph we will let m=1 and n =5 . So to satisfy both the digraph together we will have to take LCM of (3,5) which is 15
hence m=1 and n=15