The figure of a garden tile is a regular hexagon ABCDEF with a side of 2 centimetres…

Step 1. The diagonal. The centre $O$ is at distance $2$ centimetres from every vertex, so $AD = 4$ centimetres and $\overrightarrow{AO} = \dfrac{1}{2}\overrightarrow{AD}$.

Step 2. $B$ and $F$. $ABOF$ is a parallelogram with all sides $2$ centimetres. By the parallelogram law, $\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO} = \dfrac{1}{2}\overrightarrow{AD}$.

Step 3. $C$ and $E$. Let $M$ be the midpoint of $CE$. Then $\overrightarrow{AC} + \overrightarrow{AE} = 2\overrightarrow{AM}$. The point $M$ is on $OD$ and $OM = 1$ centimetre, so $AM = 3$ centimetres and $\overrightarrow{AM} = \dfrac{3}{4}\overrightarrow{AD}$. So $\overrightarrow{AC} + \overrightarrow{AE} = \dfrac{3}{2}\overrightarrow{AD}$.

Step 4. The total. $\dfrac{1}{2}\overrightarrow{AD} + \dfrac{3}{2}\overrightarrow{AD} + \overrightarrow{AD} = 3\overrightarrow{AD}$. So $k = 3$ and the length is $3 \times 4 = 12$ centimetres.