Points $R$, $S$, and $T$ lie on a segment with $RS = 6$ and $ST = 16$.

The midpoint of $\overline{RS}$ is $\frac{6}{2} = 3$ cm from $R$, so it is $3$ cm from $S$ as well.

The distance from the midpoint of $\overline{RS}$ to $T$ is:

$3 + 16 = 19 \text{ cm}$