We have an universal set U of 250 students

    and its subsets


M of 63 students;

L of 50 students;

H of unknown number of students;


   their in-pairs intersections

ML of 21 students;

MH of  8 students

LH of 5 students


    and finally their triple intersection

MHL of 1 student.


Based on this info, we can easily write an equation for the number of students who takes at least one of the three cources

    n(M U L U U) = m(M) + n(L) + n(H) - n(M ∩ L) - n(M ∩ H) - n(L ∩ H) + n(M ∩ L ∩ H). 


In this equation, the left side is 250-2 = 248.

In the right side, you know all the terms except of n(H), which I am going to find now.

Substituting all known terms, you get

    248 = 63 + 50 + n(H) - 21 - 8 - 5 + 1.


From this equation,

    n(H) = 248 - 63 - 50 + 21 + 8 + 5 - 1 = 168.


Now, to find n(H only), we should subtract n(MH) and n(LH) from n(H) and add n(MHL):

    n(H only) = 168 - 8 - 5 + 1 = 156.     ANSWER