Hi Kaushik,
It’s not easy to help you when you leave all the relevant information behind in an older thread. 
Here are summary calculations for setting up this kind of simulation:
At room conditions, the molar concentrations and molar volumes of relevant pure substances are:
Acetonitrile: 19.2 mol/L; 52.2 mL/mol (note 1)
BMIm-PF6: 4.86 mol/L; 206 mL/mol (note 1)
So as a first guess, 1.5 mol of BMIm-PF6 in 1 L would take up ~300 mL of volume, leaving ~700 mL of volume to the acetonitrile corresponding to 13.4 mol of acetonitrile.
As a nice hack, 1 mol/L corresponds, in particle concentrations, to 0.6 particles/nm^3 (note 2). Thus the earlier molar ratio (1.5 : 13.4) corresponds to 0.9 ion pairs and 8 acetonitrile molecules per nm^3, or a ratio of 0.113 ion pairs per acetonitrile molecule. That’s only 6% off from the reported value of 0.107 (230/2146), and one can quickly surmise that their molar ratio is lower because BMIm-PF6, dissolved in a polar solvent, has lower ionic strength and thus increased partial molar volume relative to the neat IL.
Hope that makes sense.
Note 1
The molar concentration of a pure substance is just the number of moles contained in 1 L of volume - so, for example, water, with molar mass 18 g/mol and room condition density 1000 g/L, has (room condition) molar concentration 1000/18 = 55 mol/L. (Thus we can consider aqueous solutions up to 2-3 M as being “dilute” – solute:solvent molar ratios under 5% – while more concentrated aqueous solutions are considerably less colligative.)
Thus, pure acetonitrile is 19.1 M (786 g/L / 41 g/mol) and pure BMIm-PF6 is 4.86 M (1380 g/L / 284 g/mol).
Note 2
1 mol / L = 6e23 particles / 1e24 nm^3 = 0.6 particle / nm^3.