First, the concentration which determines the osmostic pressure is the total concentration of ions. Assuming that the ions are completely dissociated (which is an approximation at this range of concentrations) we get a result which, while approximate, is probably in the correct ball park. Without further ado, the total concentration of ions is
$c = i_{NaOH}C_{NaOH}+ i_{MgCl_2}C_{MgCl_2}$
$c = 2(0.470) + 3(0.068)$
= 0.940 + 0.20 = 1.14 M
The temperature is 298.15K here and using R = 0.0820582, we get an osmostic pressure of
$\pi = cRT$
$= 1.14 mol/L(0.0820l.atm/mol.K)(298K)$
=27.9 atm
In this case, we have two solutions at the same temperature but at different osmotic pressures. That is,
$\pi_1V_1 = \pi_2V_2$
The volume of the more concentrated solution is, thus,
$V_2 = \frac{ pi_1V_1}{\pi_2}$
$V_2 = \frac{27.9 atm \times 1}{100 atm} = 0.279 L$
The amount of pure water produced would be just 1.00L - 0.28L = 0.72L.