College Physics (4th Edition)

The average pressure on that area is $49~atm$
We can find the pressure: $P = \frac{F}{A} = \frac{500~N}{1.0\times 10^{-4}~m^2} = 5.0\times 10^6~N/m^2$ We can convert the pressure to units of atm: $(5.0 \times 10^6~N/m^2) \times \frac{1~atm}{1.013\times 10^5~Pa} = 49~atm$ The average pressure on that area is $49~atm$