Using $13$ of the $17$ roads leaves exactly $4$ roads unused

A trail exists only if just $A$ and $L$ have odd degree; this forces which $4$ roads to drop

Name the four cities in the top row $A,B,C,D,$ those in the middle row $E,F,G,H,$ and those in the bottom row $I,J,K,L.$ A route using $13$ roads is an open Euler trail, so in the used graph exactly $A$ and $L$ have odd degree.

In the full map, the vertices whose degree parity must change are $A,B,C,E,H,J,K,L.$ Because only $4$ roads are removed, those roads must pair these $8$ vertices. Among them, $E$ is adjacent only to $A,$ forcing $AE$ to be removed; then $BC$ is forced. Similarly $H$ is adjacent only to $L,$ forcing $HL,$ and then $JK.$

The remaining graph is a chain $$\begin{gathered} A-B-F,\\ F-E-I-J-F,\\ F-G,\\ G-C-D-H-G,\\ G-K-L. \end{gathered}$$ Each of the two $4$-cycles can be traversed in either direction, and everything else is forced. Hence there are $2\cdot2=4$ routes.

Thus, E is the correct answer.