Answer will be (C) Both $I$ and $II$ are correct
For $I$ :
Lets assume that every vertex has degree 8 (Just to check whether possible or not)
By Handshaking Theorem ,
We have LHS =$\sum deg(v_{i})$ = 8*n
and RHS = 2$e$ = $2(4n-16)$=$8(n-4)$ [Now this is always true as the number of edges has been provided]
Now in order to make the LHS equal to the RHS there must be a vertex(or vertices) whose degree should be less than 8 hence $I$ is true.
For $II$
Lets take an example,
Let n=5 then No. of edges = $4n-16$=$4$ which is also equal to $n-1$
Hence it forms a tree(Suppose), then $II$ is also true as there exist a vertex such that there are less than 16 vertices at distance exactly 2 from it.