Creating Boric Acid from Borax and Hydrochloric Acid
Assuming that boric acid is completely soluble in water, you did everything correctly so far, you simply missed the last step: dilution.

First you calculated the amount of substance of boric acid you would need to produce a two molar solution. With that you basically assume, that you dissolve solid boric acid in water.
You then calculated the amount of borax you need for complete reaction with hydrochloric acid. Since this is already hydrogen chloride dissolved in water, you bring some of the water you assumed before. But this is as your calculation says only about 40&nbsp;mL.
Still assuming that it is completely soluble, you would have produced a solution that is about ten molar. You still need to dilute it to the final volume, to have the desired concentration.

Now let's go one step further. Boric acid is not very soluble in water under standard conditions, see Wikipedia. As A.K. pointed out, solubility is 47.2&nbsp;g/L at 20&nbsp;&deg;C. That translates to 0.76 mol/L. As you can see solubility increases with temperature, in boiling water the maximum concentration is about 4.5&nbsp;mol/L.
So what you are trying to attempt is not possible, at least not in pure water at room temperature. Try heating it up a bit, somewhere around 70&nbsp;&deg;C should be enough. 

Alternatively PubChem has kindly provided us with a little more statistics about solubility.
In methanol you can obtain the desired two molar solution at room temperature, as solubility is 173.9&nbsp;g/L at 25&nbsp;&deg;C.