What is the answer? 2+3*(1680/2 -1)

The answer is $2016$.

Could you tell me how to solve this problem???Thx

We claim that the answer at least $\lceil {\frac{6}{5}n}\rceil$. It is tight for $n=5k$. To see this, first construct a cycle of length $4k$, and number the elements $1,2,\ldots 4k$. Then, we create $k$ new points $P_i$, each of which are connected to $i$ and $2k+i$. The number of degrees here is original $4k$ + new $2k$ = $6k$. To verify that this construction works, note that if we cut $<2$ from the $4k$-cycle, everything is connected to the $4k$-cycle still. Otherwise, if we cut $2$ from the outer ring, you'll always be able to use the paths from $i\to P_i\to 2i+1$ to make your way anywhere. To now prove this is optimal, we use induction. Case 1: There exists a pair of linked vertices $A,B$ where $\deg A = \deg B = 2$. Then, we must have that $A,B$ are both connected to a third $C$. Thus, by the inductive hypothesis on everything except for $A,B$. \[f(n) \geq 3 + f(n-2) > \lceil \frac{6n}{5}\rceil\]Now, for the second case, no two elements with degree 2 are incident. Say that there are $x$ with degree 2, and $n-x$ vertices with degree $\geq 3$. Then, we firstly have that the set of edges from $x$ are disjoint, so \[E\geq 2x\]and since the average degree of all $n-x$ are $\geq 3$, by handshake lemma we have \[E\geq \frac{2x+3(n-x)}{2}=\frac{3n-x}{2}\]Combining, $5E = E+4E\geq 2x + 4\cdot \frac{3n-x}{2} = 6n$. Thus, $E\geq \frac{6n}{5}$ and we're done. $\blacksquare$. To finish, we now just compute \[\frac65 \cdot 1680 = 2016\]