CN-TST 2011 Problem 15

Let $G$ be a simple graph with $3n^2$ vertices ( $n\geq 2$ ). It is known that the degree of each vertex of $G$ is not greater than $4n$ , there exists at least a vertex of degree one, and between any two vertices, there is a path of length $\leq 3$ . Prove that the minimum number of edges that $G$ might have is equal to $\frac{(7n^2- 3n)}{2}$ .