Answer
$$d_s=50\;cm$$
Work Step by Step
Let $m$ be the mass of the child.
When a child is placed in the seat, the both cords descend by a distance $d_s$ as the cords stretch. Treating the cords as springs, can write
$$kd_s=mg$$
where $k$ is the spring constant.
Therefore,
$$\boxed{k=\frac{mg}{d_s}}$$
When seat is pulled down an extra distance $d_m$ and released, so that the child oscillates vertically. For the vertical oscillation of the child, we can wite
$$F=kd_m$$
$$or,\;ma=kd_m$$
$$or,\;a=\frac{mg}{d_s}\times \frac{d_m}{m}$$
$$or,\;0.2g=\frac{d_mg}{d_s}$$
$$or,\;d_s=\frac{d_m}{0.2}$$
Substituting the given value of $d_m$, we obtain
$$or,\;d_s=\frac{10}{0.2}\;cm$$
$$or,\;\boxed{d_s=50\;cm}$$
