Denote the vertices of the dodecagon as on Fig. 21. Since $QI = AS$ and $IS = QA$, we have
$$DQ + QI + IS + SG = DQ + AS + QA + SG = DA + AG.$$
Similarly we obtain $ER + RK + KP + PB = EA + AB$. Therefore
$$\begin{align*} & CD + DQ + QI + IS + SG + GF + FE + ER + RK + KP + PB + BC \\ &= CD + DA + AG + GF + FE + EA + AB + BC \\ &= AB + BC + CD + DA + AE + EF + FG + GA. \end{align*}$$
So the sum of the perimeters of the squares $ABCD$ and $AEFG$ is equal to the perimeter of the dodecagon (72 cm).

Since $QA = IS$, $AP = RK$, $SA = IQ$ and $AR = PK$, we have
$$\begin{aligned} & AP + PH + HQ + QA + AR + RJ + JS + SA \\ &= RK + PH + HQ + IS + PK + RJ + JS + IQ \\ &= HI + IJ + JK + KH. \end{aligned}$$
So the perimeter of the square $HIJK$ is equal to the sum of the perimeters of the rectangles $APHQ$ and $ARJS$. However we are given that the sum of the perimeters of $APHQ$ and $ARJS$ is equal to one fifth of the sum of the perimeter of $ABCD$, the perimeter of $AEFG$ and twice the perimeter of $HIJK$.
Denoting the side length of $HIJK$ by $x$ cm, its perimeter will be $4x$ cm. Then we can combine the relations of the previous two paragraphs into the equation
$$4x = \frac{1}{5} (72 + 2 \cdot 4x).$$
Solving it, we obtain $4x = 24$, which gives $x = 6$. So the side length of the square $HIJK$ is 6 cm.