Given,
$k \cdot 10^{9}=10^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\ldots+10(11)^{9} $
$\Rightarrow k=1+2\left(\frac{11}{10}\right)+3\left(\frac{11}{10}\right)^{2}+\ldots+10\left(\frac{11}{10}\right)^{9} . \ldots$ (i)
$\left(\frac{11}{10}\right) k=1\left(\frac{11}{10}\right)+2\left(\frac{11}{10}\right)^{2}+\ldots+9\left(\frac{11}{10}\right)^{9}+10\left(\frac{11}{10}\right)^{10} \ldots $ (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$k\left(1-\frac{11}{10}\right) =1+\frac{11}{10}+\left(\frac{11}{10}\right)^{2}+\ldots+\left(\frac{11}{10}\right)^{9}-10\left(\frac{11}{10}\right)^{10}$
$\Rightarrow k\left(\frac{10-11}{10}\right) =\frac{1\left[\left(\frac{11}{10}\right)^{10}-1\right]}{\left(\frac{11}{10}-1\right)}-10\left(\frac{11}{10}\right)^{10}$
$\left[\because \text{In} GP\right.$, sum of $n$ terms $=\frac{a\left(r^{n}-1\right)}{r-1}$, when $\left.r>1\right]$
$\Rightarrow -k=10\left[10\left(\frac{11}{10}\right)^{10}-10-10\left(\frac{11}{10}\right)^{10}\right]$
$\therefore k=100$