If $$\sum_{r=0}^{10} \left(\frac{10^{r+1} - 1}{10^r}\right) \cdot \,^{11}C_{r+1} = \frac{\alpha^{11} - 11^{11}}{10^{10}}$$, then $$\alpha$$ is equal to :

The given sum is

$$S=\sum_{r=0}^{10}\left( \frac{10^{\,r+1}-1}{10^{\,r}} \right)\,{}^{11}C_{\,r+1}\,.\tag{-1}$$

Simplify the fraction inside the summation:

$$\frac{10^{\,r+1}-1}{10^{\,r}} =\frac{10^{\,r+1}}{10^{\,r}}-\frac{1}{10^{\,r}} =10-10^{-\,r}\,.\tag{-2}$$

Using $$( -2 )$$ in $$( -1 )$$ gives

$$S=\sum_{r=0}^{10}\Bigl(10-10^{-\,r}\Bigr)\,{}^{11}C_{\,r+1} =10\sum_{r=0}^{10}{}^{11}C_{\,r+1}-\sum_{r=0}^{10}10^{-\,r}\,{}^{11}C_{\,r+1}\,.\tag{-3}$$

Put $$k=r+1$$ so that $$k$$ runs from $$1$$ to $$11$$:

$$S=10\sum_{k=1}^{11}{}^{11}C_{\,k}-\sum_{k=1}^{11}10^{-(k-1)}\,{}^{11}C_{\,k}\,.\tag{-4}$$

First summation
Using $$\sum_{k=0}^{11}{}^{11}C_{\,k}=2^{11}$$, we get

$$\sum_{k=1}^{11}{}^{11}C_{\,k}=2^{11}-{}^{11}C_{\,0}=2^{11}-1\,.$$

Hence

$$10\sum_{k=1}^{11}{}^{11}C_{\,k}=10\left(2^{11}-1\right)\,.\tag{-5}$$

Second summation
Write $$10^{-(k-1)}=10\cdot10^{-k}$$, so

$$\sum_{k=1}^{11}10^{-(k-1)}\,{}^{11}C_{\,k} =10\sum_{k=1}^{11}10^{-k}\,{}^{11}C_{\,k}\,.$$

Using the binomial expansion $$(1+x)^{11}=\sum_{k=0}^{11}{}^{11}C_{\,k}x^{k}$$ with $$x=\tfrac1{10}$$:

$$\sum_{k=1}^{11}{}^{11}C_{\,k}10^{-k}=(1+\tfrac1{10})^{11}-1 =\left(\tfrac{11}{10}\right)^{11}-1\,.$$

Therefore

$$\sum_{k=1}^{11}10^{-(k-1)}\,{}^{11}C_{\,k} =10\Bigl[\left(\tfrac{11}{10}\right)^{11}-1\Bigr] =\frac{11^{11}-10^{11}}{10^{10}}\,.\tag{-6}$$

Substituting $$( -5 )$$ and $$( -6 )$$ into $$( -4 )$$:

$$S=10\left(2^{11}-1\right)-\frac{11^{11}-10^{11}}{10^{10}} =20470-\frac{11^{11}-10^{11}}{10^{10}}\,.\tag{-7}$$

The question states that

$$S=\frac{\alpha^{11}-11^{11}}{10^{10}}\,.\tag{-8}$$

Equate $$( -7 )$$ and $$( -8 )$$, then multiply by $$10^{10}$$:

$$20470\cdot10^{10}-\bigl(11^{11}-10^{11}\bigr)=\alpha^{11}-11^{11}\,.$$

Rearrange:

$$\alpha^{11}=20470\cdot10^{10}+10^{11}\,.$$

Factor out $$10^{10}$$:

$$\alpha^{11}=10^{10}\left(20470+10\right)=10^{10}\cdot20480\,.$$

Notice that $$2048=2^{11}$$, so

$$20480=2048\cdot10=2^{11}\cdot10\,.$$

Hence

$$\alpha^{11}=10^{10}\cdot2^{11}\cdot10 =\bigl(10\cdot2\bigr)^{11}=20^{11}\,.$$

Taking the positive 11-th root (since $$\alpha$$ is positive), we get

$$\alpha=20\,.$$

Therefore, Option D is correct.