Spoiler:  Inclusion Exclusion Principle says \[|A \cup B \cup C| = |A| + |B| + |C| – |A \cap B| – |B \cap C| – |B \cap C| + |A \cap B \cap C|.\]   Since there are 20 people in each of $A$, $B$, $C$ individually, the only possible number given in the table for the total number of people is $34$.   But now we have \[34 = 20 + 20 + 20 – 10 – 10 – 10 + |A \cap B \cap C|\] which gives $|A \cap B \cap C| = 4.$

My mathematician friend Peter Ross from Santa Clara College in California comments:

Inclusion-exclusion isn’t needed, given that the answers must be among the 5 given. In fact, using two 3-set Venn Diagrams, one with 4 in the triple intersection and one with 8 in it (the only possible values since the double intersections have only 10), just fill out uniquely the other regions. The first case yields the answers 4 and 34 total, while the second case gives 8 and 38 total, so it’s excluded and the answers are 4 and 34. I imagine it’s called a two-part problem as there are two cases.  Moral–think visually not symbolically!

To which I add:

Algebra can play a role too! The 3 intersecting sets Venn diagram has 8 regions, one of which [the complement of the triple union] we are told has no elements in it. Filling in the rest using $t$ for the count in the triple intersection, $d$ for the (necessarily identical) double (only) intersection counts and $s$ for the (necessarily identical) counts of the elements in just one single set, we get $t + d = 10$ for each double intersection, and $t + 2d + s = 20$ for each individual set.

Subtracting, we see that $d + s = 10$ also, and comparing with the above we deduce that $t = s$.

Now use the fact of the choices on offer in the table, the total number of people (which is at least 20) is now $4t + 3d$ and must equal 34. Since $t + d = 10$ we also have $3t + 3d = 30$, and subtracting yields $t = 4$ as desired.