If $$\sum_{i=1}^{20} \left(\frac{^{20}C_{i-1}}{^{20}C_i + ^{20}C_{i-1}}\right)^3 = \frac{k}{21}$$, then $$k$$ equals:

We begin with the expression

$$S \;=\;\sum_{i=1}^{20}\left(\dfrac{{}^{20}C_{\,i-1}}{{}^{20}C_{\,i}+{}^{20}C_{\,i-1}}\right)^{\!3}$$

For each index $$i$$ we isolate the inner fraction

$$T_i \;=\;\dfrac{{}^{20}C_{\,i-1}}{{}^{20}C_{\,i}+{}^{20}C_{\,i-1}}.$$

To simplify it, we recall the relation connecting two successive binomial coefficients

$$^{20}C_i \;=\;^{20}C_{\,i-1}\;\times\;\dfrac{20-(i-1)}{i}\;=\;^{20}C_{\,i-1}\;\times\;\dfrac{21-i}{i}.$$

Substituting this value in the denominator of $$T_i$$, we get

$$^{20}C_{\,i}+{}^{20}C_{\,i-1} \;=\;^{20}C_{\,i-1}\left(\dfrac{21-i}{i}+1\right) \;=\;^{20}C_{\,i-1}\left(\dfrac{21-i+i}{i}\right) \;=\;^{20}C_{\,i-1}\left(\dfrac{21}{i}\right).$$

Hence

$$T_i \;=\;\dfrac{^{20}C_{\,i-1}}{^{20}C_{\,i-1}\left(\dfrac{21}{i}\right)} \;=\;\dfrac{1}{\dfrac{21}{i}} \;=\;\dfrac{i}{21}.$$

We need the cube of this term, so

$$T_i^{\,3}\;=\;\left(\dfrac{i}{21}\right)^{\!3}\;=\;\dfrac{i^{3}}{21^{3}}.$$

Substituting back into the summation, we obtain

$$S \;=\;\sum_{i=1}^{20}\dfrac{i^{3}}{21^{3}} \;=\;\dfrac{1}{21^{3}}\sum_{i=1}^{20} i^{3}.$$

Now we state the standard formula for the sum of cubes:

$$\sum_{i=1}^{n} i^{3} \;=\;\left[\dfrac{n(n+1)}{2}\right]^{2}.$$

Putting $$n=20$$, we have

$$\sum_{i=1}^{20} i^{3} \;=\;\left[\dfrac{20\times 21}{2}\right]^{2} \;=\;(10\times 21)^{2} \;=\;210^{2} \;=\;44100.$$

Therefore

$$S \;=\;\dfrac{1}{21^{3}}\times 44100.$$

Since $$21^{3}=21\times 21^{2}=21\times 441=9261$$, we get

$$S \;=\;\dfrac{44100}{9261}.$$

Factorising the numerator, note that $$44100=441\times 100$$, and because $$9261=21\times 441$$, we can write

$$S \;=\;\dfrac{441\times 100}{21\times 441} \;=\;\dfrac{100}{21}.$$

The question states $$S=\dfrac{k}{21}$$, so by equating we find

$$k \;=\;100.$$

Hence, the correct answer is Option B.