#### Answer

The air pressure at the floor is greater than the air pressure at the ceiling by 0.20%

#### Work Step by Step

The pressure at the floor of the gym is $1 ~atm$. The pressure at the gym floor is greater than the pressure at a height $h$ by $\rho~g~h$. Let $P_f$ be the pressure at the floor. Let $P_c$ be the pressure at a height of 16 meters.
$P_c+\rho~g~h = P_f$
$P_c = P_f - \rho~g~h$
$P_c = (1.013\times 10^5~N/m^2)-(1.29~kg/m^3)(9.80~m/s^2)(16~m)$
$P_c = 1.011\times 10^5~N/m^2$
We can find the percent increase of the air pressure at the floor compared to the air pressure at the ceiling.
$\frac{P_f-P_c}{P_c}\times 100\% = \frac{(1.013\times 10^5~N/m^2)-(1.011\times 10^5~N/m^2)}{1.011\times 10^5~N/m^2} \times 100\% = 0.20\%$
The air pressure at the floor is greater than the air pressure at the ceiling by 0.20%.