First lets' state the Universal Gas Constant: R = (0.08206 x L x atm) / (mol x K)
And the Ideal Gas Law: PV = nRT
Now, lets find a handy calculator to do everything for us: http://www.ajdesigner.com/idealgas/ideal_gas_law_mole_equation.php
Now we can rearrange the gas law to solve for moles: n = PV/RT... and do some conversions... 320 °C = 593.15 K, and 1600 psig = 108.87 atmospheres (or did you mean psia?)... or, just plug the numbers into the calculator.
So, if the pressure is 1600 psi at 320 °C in a 20 L vessel you need 44.735912618737 moles of gas. Now, we multiply by the molar mass of water (18.01528 g/mol) to obtain more familiar values... it appears that you will need 805.93 grams of water (more or less). And let's use the density of water at room temperature to find the volume... 805.93g / 998.23 g/L = 0.80736 liters.
My advice, for what it's worth, is- if you are actually going to do this, you should probably use more water, and let the steam boil for a few minutes to evacuate the air from the vessel before locking it shut. Oh, and be sure to wear eye protection while you run for cover.
Edit based on comment:
Question: ...for how long would the pressure be mainted if [water vapor] is exiting also? I mean, I will start release pressure as soon as it hits 1600PSI, but I'll keep on heating the water too. So, for how long will I be able to maintain that 1600PSI exit force?
2nd Answer: The loss of pressure would be instantaneous... concurrent with the loss of water. To maintain the pressure you would need to increase the temperature as the water exits. The time it would last, as it empties and as the temperature increases, depends on the rate/amount of water being released and/or the rate of temperature increase. If you want to maintain 1600 psi, then you either need more water or more heat; if you use more water, then you could maintain the 320 °C because you would no longer have 20 L of "head-space" in the vessel. The amount of vapor in the head-space of the vessel would be less than 805.93g. Not that it matters, but to calulate the amount of vapor, you would subtract the liquid volume from the total vessel volume because water is practically incompressible (maybe around 1% at 320 °C). The way to maintain 1600 psi would be to fill the tank (maybe 2/3 full), and use a pressure relief valve (like all pressure cookers do). A pressure relief valve would automatically control the pressure regardless of the change in temperature and volume (as long as it's sufficient or in excess- to supply the minimum amount of pressure to the valve).
Supposing that you had an infinite amount of heat on tap, the pressure would (theoretically) be maintained until the last molecule of water escaped the vessel. Now, let's forget that steel melts at about 1370 °C and water actually breaks down at around 2500 °C (says google- not sure if that chages because pressure would be 1600 psi) and steel burns at about 4000 °C; releasing one molecule of water per hour, the pressure would be maintained until you depleated 44.736 moles (so, multiplying by Avagadro's number: 6.022×10^23), you will have 2.694×10^25 hours.
To find the necessary temperature for 1 molecule of H2O to supply 1600 psi in 20L (the final temperature at hour 2.694×10^25), convert 1 molecule to moles and rearrange the gas law to solve for T.
n = 6.022×10^-23 : T = PV/Rn
T(final) = (108.87atm x20L)/(0.08206 x6.022×10^-23mol) = (about) 4.406×10^26 K. To convert K to °C, subtract 273.15 from 440,600,000,000,000,000,000,000,000 K.