Answer
$1.81\times10^{-2}m^{3}/s$
Work Step by Step
Let's apply the equation $A_{1}v_{1}=A_{2}v_{2}$ to get a combination between speeds.
$A_{1}v_{1}=A_{2}v_{2}$
$\pi (0.02\space m)^{2}v_{1}=\pi (0.04)^{2}v_{2} $
$v_{1}=4v_{2}-(1)$
Let's apply Bernoulli's equation to find $v_{2}$
$P+\frac{1}{2}\rho v^{2}+\rho gh=constant$
$P+\frac{1}{2}\rho v_{1}^{2}+0=P+\frac{1}{2}\rho v_{2}^{2}+\rho gh$
$v_{1}^{2}-v_{2}^{2}=2gh-(2)$
(1)=>(2),
$15v_{2}^{2}=2gh=>v_{2}=\sqrt {\frac{2gh}{15}}$
$v_{2}=\sqrt {\frac{2(9.8\space m/s)(10\space m)}{15}}=3.61\space m/s$
Volume flow rate = Av = $\pi (0.04\space m)^{2}(3.61\space m/s)=1.81\times10^{-2}m^{3}/s$
