Every new set $S_{n+1}$ starts after $n$ elements from the starting element of $S_{n}$ . This means that we can find the starting number of $S_{21}$ using Arithmetic progression formula.
Let $\text{Sum}\left({n}\right)$denote sum of natural numbers until ${n:}$
$S_1$ starts with $= 1$
$S_2$ starts with $= \text{Sum}\left(1\right) + 1= 2$
$S_3$ starts with $= \text{Sum}\left(2\right) + 1 = \left(1+2\right) + 1 = 4$
$\text{Similarly, }S_{21}$ starts with $S\left(20\right) +1 = \frac{20(20+1)}{2} + 1 = 211$
Now we need to find sum of ${21}$ consecutive natural numbers starting from ${211}$
Using A.P. sum formula $S_n = \frac{n}{2} \left[2a + \left(n-1\right)*d \right]$ where, $a=$ starting term , $d=$ difference
Sum of elements in $S_{21} = \frac{21}{2}\left[2\left(211\right) + 20 \right] = \frac{21 * 442}{2} = 4641.$ Option (D) is correct