$\begin{array}{1 1}(a)\;0.01M&(b)\;0.001M\\(c)\;0.1M& (d)\;0.02M\end{array} $

$6.023\times 10^{23}$ molecule of urea=1mol

$6.0\times 10^{20}$ molecule of urea =$\large\frac{6.0\times 10^{20}}{6.023\times 10^{23}}$$=10^{-3}$ mol

$\therefore 10^{-3}$ mol of urea are present in 100 ml

$\therefore$ moles of urea present in 1000ml=$10^{-3}\times \large\frac{1000}{100}$$=10^{-2}$

$\therefore$ molarity of solution =$10^{-2}M=0.01$M

Hence (a) is the correct option.

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