Solution:

We case on the value of friends that each person has. This value ranges from $1$ to $4$, since the graph is neither empty nor complete.

Note that the cases for $1$ and $2$ friends correspond with the case for $4$ and $3$ friends, since choosing who are friends determines who are not friends.

Case 1: everyone has $1$ friend

This means that the $6$ people must split up into $3$ pairs where the people in each pair are friends.

There are $5$ choices for the friend for the first person. This leaves $4$ people remaining.

There are then $3$ choices for the friend of the next unpaired person. The remaining $2$ people are then forced to be friends.

Therefore, there are $3 \cdot 5 = 15$ possibilities for this case.

Case 2: everyone has $2$ friends

There are two possibilities for this case. There could be two triples where everyone in a triple is friends with each other.

For this possibility, there are $\binom{6}{3} = 20$ ways to choose the people in the first triple. We have to divide by $2$ since we can swap the pairs. This gives us $20 \div 2 = 10$ configurations.

The second possibility is if the friends form a hexagon where the person at each vertex is friends with the adjacent people.

The first person can be placed anywhere on the hexagon. There are $\binom{5}{2} = 10$ ways to choose the people adjacent to this person.

The final $3$ people can be placed in $3! = 6$ ways in the remaining spots. This case then has a total number of $$10 + 10 \cdot 6 = 70$$ configurations.

The total number of arrangements is then $$2(15 + 70) = 170.$$

Thus, B is the correct answer.