Answer
$0.05\space m^{3}/s$
Work Step by Step
Let's apply Bernoulli's equation to find the speed of the water entering the hold,
$P+\frac{1}{2}\rho v^{2}+\rho gh=constant$
$P+\frac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P+\rho gh_{2}$
$v_{1}=\sqrt {2g(h_{2}-h_{1})}$ ; Let's plug known values into this equation.
$v_{1}=\sqrt {2(9.8\space m/s^{2})(2\space m)}=6.26\space m/s$
We know that,
Volume flow rate (Q)= Area times speed (Av)
$Q=(8\times10^{-3}m^{2})(6.26\space m/s)=0.05\space m^{3}/s$
