We must ensure that any pair of students enrolled in the same course has the same university.
So choose arbitrary enrolled students $s_1$ and $s_2$.
If both are enrolled in course $c$, then require $u_1=u_2$.
Hence,
$(\mathrm{Student}(s_1,n_1,u_1)\land \mathrm{Enrollment}(s_1,c)\land \mathrm{Student}(s_2,n_2,u_2)\land \mathrm{Enrollment}(s_2,c))\Rightarrow u_1=u_2$.
This condition must hold for all pairs, so universal quantification is required.
A useful equivalent form is, there must not exist two students enrolled in the course with $u_1\neq u_2$.
That follows from
$\forall x(P(x)\Rightarrow Q(x))\equiv\neg\exists x(P(x)\land\neg Q(x))$.
Let's see what happens for a course with zero or one enrolled student.
There is no pair violating the condition, so the universal expression is true.
Therefore such courses are correctly retained.
Hence,
Answer : $\boxed{\mathrm{C}}$
