Use the identity: $\dfrac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}} = \dfrac{{}^{20}C_{i-1}}{{}^{21}C_i} = \dfrac{i}{21}$

So the sum becomes: $\displaystyle\sum_{i=1}^{20}\left(\frac{i}{21}\right)^3 = \frac{1}{21^3}\sum_{i=1}^{20}i^3 = \frac{1}{21^3}\cdot\left(\frac{20\cdot21}{2}\right)^2 = \frac{(210)^2}{21^3} = \frac{44100}{9261} = \frac{200}{21}$

Therefore $k = \mathbf{200}$.