Answer: B

Solution:

Use $F_{2k} = F_k L_k,$ so each term $\dfrac{F_{2k}}{F_k} = L_k,$ the $k$th Lucas number. That collapses the sum to $\sum_{k=1}^{10} L_k.$ With $L_1 = 1,$ $L_2 = 3,$ $L_3 = 4,$ $\ldots,$ $L_{10} = 123,$ the identity $\sum_{k=1}^{n} L_k = L_{n+2} - 3$ gives $L_{12} - 3 = 322 - 3 = 319.$ Thus, B is the correct answer.