1. Rewrite the summand

Given $\sum_{r=0}^{10} \left(\frac{10^{r+1}-1}{10^r}\right)\binom{11}{r+1}$

first simplify the factor: $\frac{10^{r+1}-1}{10^r}=10-10^{-r}.$

So the sum becomes $S=\sum_{r=0}^{10}\left(10-\frac1{10^r}\right)\binom{11}{r+1}.$

Let $k=r+1.$ Then as $r$ goes from $0$ to $10$, $k$ goes from $1$ to $11$. Also, $r=k-1$. Hence $S=\sum_{k=1}^{11}\left(10-\frac1{10^{k-1}}\right)\binom{11}{k}.$

Now split the sum: $S=10\sum_{k=1}^{11}\binom{11}{k}-\sum_{k=1}^{11}\frac{1}{10^{k-1}}\binom{11}{k}.$

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2. Evaluate the first sum

Using $\sum_{k=0}^{11}\binom{11}{k}=2^{11}=2048,$ we get $\sum_{k=1}^{11}\binom{11}{k}=2^{11}-1=2047.$

Therefore, $10\sum_{k=1}^{11}\binom{11}{k}=10(2047)=20470.$

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3. Evaluate the second sum using the binomial theorem

Consider $\sum_{k=0}^{11}\binom{11}{k}x^k=(1+x)^{11}.$

Take $x=\frac{1}{10}$: $\sum_{k=0}^{11}\binom{11}{k}\left(\frac1{10}\right)^k=\left(1+\frac1{10}\right)^{11}=\left(\frac{11}{10}\right)^{11}.$

We need $\sum_{k=1}^{11}\frac{1}{10^{k-1}}\binom{11}{k}=10\sum_{k=1}^{11}\binom{11}{k}\left(\frac1{10}\right)^k.$

So,

=10\left[\left(\frac{11}{10}\right)^{11}-1\right].$$ --- 4. **Substitute back into $S$** Thus, $$S=20470-10\left[\left(\frac{11}{10}\right)^{11}-1\right].$$ Simplify: $$S=20470-10\left(\frac{11^{11}}{10^{11}}-1\right) =20470-\frac{11^{11}}{10^{10}}+10.$$ Hence $$S=20480-\frac{11^{11}}{10^{10}}.$$ Write $20480$ with denominator $10^{10}$: $$20480=\frac{20480\cdot 10^{10}}{10^{10}}.$$ Therefore, $$S=\frac{20480\cdot 10^{10}-11^{11}}{10^{10}}.$$ Comparing with $$S=\frac{\alpha\cdot 11-11^{11}}{10^{10}},$$ we get $$11\alpha=20480\cdot 10^{10}.$$ This comparison clearly suggests the printed expression is intended as $$\frac{\alpha-11^{11}}{10^{10}},$$ which is the standard form matching our simplification. Then $$\alpha=20480\cdot 10^{10}=2048\cdot 10^{11}=2^{11}\cdot 10^{11}=20^{11}.$$ So, $$\alpha=20^{11}.$$ Since the options are ordinary numbers, the intended question is asking for the base, i.e. $\alpha=20$. --- 5. **Check the options** Thus the correct option is: $$\boxed{20}$$ which is **Option B**.