one mole of a gas occupies 22.4 L at 0°C and 760 mm Hg. ?

(22.4 L/mol) x (2.00 mol) = 44.8 L

(760 mmHg) x (44.8 L / 3.00 L) x (25 + 273) K / (273 K) = 12389 mmHg =

12400 mmHg

[Kenny B: Knowing the molar volume relieves you from knowing any value for R.]

assuming.... the gas behaves ideally... (which is not usually a good assumption at pressures > 10 atm)

.. PV = nRT

.. P = nRT/V

.. P = (2.00mol) * (0.08206 Latm/molK) * (298.15K) / (3.00L) = 16.3atm

or if you want that in mmHg

.. P = (2.00mol) * (0.08206 Latm/molK) * (298.15K) / (3.00L) * (760mmHg / 1atm)

.. .. = 1.24x10^4 mmHg

************

alternately... (still assuming ideal gas behavior)

.. P1V1 / (n1T1) = P2V2 / (n2T2)

.. P2 = P1 * (V1 / V2) * (T2 / T1) * (n2 / n1)

.. P2 = 760mmHg * (22.4L / 3.00L) * (298.15K / 273.15K) * (2 mol / 1 mol)

.. .. .. = 1.24x10^4 mmHg

same answer

************

under elevated pressures, gases don't behave ideally.  Each gas is different but they ALL deviate from ideal gas behavior.  Take ammonia for example.  via the VDW equation for real gases

.. (P + an²/V²) * (V - nb) = nRT --> P = nRT/(V-nb) - an²/V²

and from here https://en.wikipedia.org/wiki/Van_der_Waals_consta...

.. a = 4.225 L² bar / mol² * (1atm/1.01325 bar) = 4.170 L² atm / mol²

.. b = 0.0371 L/mol

so that

.. P = 14.9 atm = 1.13x10^4 mmHg

and our "ideal gas" estimate was off by almost 10%

show this to your teacher. 

Use the equation

P(1)V(1)/T(1) = P(2)V(2)/T(2)

Hence P(2) = P(1)V(1)T(2) / (T(1)V(2) 

P(2) = 760 X22,4 X 298K / 3.00 x 273 

P(2) = 6194.32 mm Hg 

This is approximately 8.15 atms. 

Use the ideal gas equation:  PV = nRT and solve for P

P = nRT/V = 2*(0.08206*298)/3.  You answer will be in atm.

I don't see that you need the volume of gas at STP to solve this question.