Answer should be option D.
Contradiction for A. Let $L_2 = \{a,b\}^*$ ... which is regular.
And $L_1 = a^nb^n$ which is CFL but not regular.
And here $L_1$ is subset of $L_2$.
Contradiction for B. Let $L_1 = ab$ ,
which is regular. And $L_2 = a^nb^n$ which is CFL but not regular.
And here $L_1$ is subset of $L_2$.
C $\rightarrow$ False, ( reason A and B).