$6.023\times 10^{23}$ 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.