The perimeter of the large triangle is $24 \; \text{cm}$, so it has side of length $8 \; \text{cm}$, so each small triangle has side of length $1 \; \text{cm}$. So each small triangle has perimeter $3 \; \text{cm}$

\begin{eqnarray}\text{Total length of black lines} &=& \text{Total perimeter of shaded triangles} \\ &=& \text{Number of shaded triangles} \times 3\; \text{cm} \\ &=& 27 \times 3\; \text{cm}\\ &=& 81\; \text{cm} \end{eqnarray}

*This problem is taken from the UKMT Mathematical Challenges.*