Momentum imparted to the surface in one collision,

$$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$$ ..... (i)

Force on the surface due to n collision per second, $$F = {n \over t}(\Delta p) = n\Delta p$$ ($$\because$$ $$t = 1s$$)

= 2 mnv [from Eq. (i)]

So, pressure on the surface,

$$p = {F \over A} = {{2mnv} \over A}$$

Here, m = 10^$$-$$26 kg, n = 10²² s^$$-$$1, v = 10⁴ ms^$$-$$1, A = 1 m²

$$\therefore$$ Pressure, $$p = {{2 \times {{10}^{ - 26}} \times {{10}^{22}} \times {{10}^4}} \over 1} = 2$$ N/m²

So, pressure exerted is of order of 10⁰.