Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).
At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).
We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).