The chemical reaction of this statement is,
$6 {\mathit{CO}}_2 + 6 H_2\mathit{O }\to C_6{H}_{12}{O}_6+ 6O_2$

1 mole of glucose needs 6 moles of water
$\mathrm{or} 180 \mathrm{g} \mathrm{of} \mathrm{glucose} \mathrm{needs} (6\mathrm{ }\times \mathrm{ }18) \mathrm{g} \mathrm{of} \mathrm{water}$
$\mathrm{or} 1 \mathrm{g} \mathrm{of} \mathrm{glucose} \mathrm{needs}=\frac{108}{180}$
$\mathrm{or} 18 \mathrm{g} \mathrm{of} \mathrm{glucose} \mathrm{would} \mathrm{need} =\mathrm{ }\frac{108\mathrm{ }\times 18}{180}$
                                                          $\Rightarrow$ 10.8 g
So, 10.8 g of water would be required to produce 18 g of glucose.