1 mole at those conditions takes up 0.123 l. Why? If 16,136.7 moles of hydrogen take up 2,000 l, then 1 mole will take up 0,123 l. You can also apply the ideal gas equation to check it:
$V=n*R*T/p=1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.123 [l]$
And what about reality? Well, ideal gas equation fails at such high pressures because it doesn't regard the space occupied by the mollecules of the gas (important at those high pressures becuase collisions are more frequent and repulsion forces more intense, therefore there's an increase in volume). Note: This is true for hydrogen, but not for all real gases.
Let's correct the ideal gas equation with the compressibility factor $z$:
Now we'll have $p*V=z*n*R*T$
At 20 MPa (200 bar) and 25 ºC we have $z=1.124$ (mean value between 1.0601 and 1.1879, highlighted down here):
Source:Hydrogen compressibility at different temperatures and pressures
So let's do the math:
$V=z*n*R*T/p=1.124*1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.139 [l]$
And 2,246.7 l if you consider the 16,137.7 moles. So prepare a bigger tank for your hydrogen!