Answer
Air enters at B and exits at A
Work Step by Step
Here we use Bernoulli's equation as follows.
$P+\frac{1}{2}\rho V^{2}+\rho gh= constant$
$P_{B}+\frac{1}{2}\rho V_{1}^{2}+0=P_{A}++\frac{1}{2}\rho V_{2}^{2}+0$
$P_{B}-P_{A}=\frac{1}{2}\rho (V_{2}^{2}-V_{1}^{2})=\frac{1}{2}\times1.29\space kg/m^{3}[(8.5\space m/s)^{2}-(1.1\space m/s)^{2}]$
$P_{B}-P_{A}=46\space Pa$
We know that the air flows from high pressure to low pressure. So it enters at B and exits at A
