1. Given data

• Mass of one hydrogen molecule: $m = 3.32\times 10^{-27}\ \text{kg}$
• Number of molecules striking per second: $n = 10^{23}\ \text{s}^{-1}$
• Area of wall: $A = 2\ \text{cm}^2 = 2\times 10^{-4}\ \text{m}^2$
• Speed of rebound (and hence incident speed, since collision is elastic): $v = 10^3\ \text{m s}^{-1}$
• Angle with the normal: $\theta = 45^\circ$

2. Velocity component normal to the wall

Only the normal component contributes to pressure: $v_n = v\cos\theta = 10^3\cos45^\circ = \frac{10^3}{\sqrt{2}}$

3. Change in momentum for one molecule

Since the molecule rebounds elastically, the normal component reverses direction.

So change in momentum per molecule is: $\Delta p = 2mv_n = 2m v\cos\theta$

Substitute values: $\Delta p = 2\times 3.32\times 10^{-27}\times 10^3\times \frac{1}{\sqrt{2}}$

$\Delta p = 6.64\times 10^{-24}\times \frac{1}{\sqrt{2}}$

$\Delta p \approx 4.70\times 10^{-24}\ \text{kg m s}^{-1}$

4. Force on the wall

If $10^{23}$ molecules strike per second, then force is rate of change of momentum: $F = n\Delta p = 10^{23}\times 4.70\times 10^{-24}$

$F = 4.70\times 10^{-1}\ \text{N} = 0.47\ \text{N}$

5. Pressure on the wall

$P = \frac{F}{A} = \frac{0.47}{2\times 10^{-4}}$

$P = 2.35\times 10^3\ \text{N m}^{-2}$

6. Match with options

This corresponds to: $\boxed{2.35\times 10^3\ \text{N m}^{-2}}$

So the correct option is A.