Answer
$1.5\times 10^{-5}\frac{m^3}{s}$
Work Step by Step
According to equation of continuity
$A_1V_1=A_2V_2$
$\implies \frac{\pi D_1^2V_1}{4}=\frac{\pi D_2^2V_2}{4}$
This can be rearranged as:
$V_1=V_2(\frac{D_2}{D_1})^2$
We plug in the known values to obtain:
$V_1=V_2(\frac{0.69}{1.9})^2=0.113V_2$
$\implies V_2^2-(0.113V_2)^2=0.225$
$V_2=\frac{\sqrt{0.225}}{1-0.113^2}=0.477\frac{m}{s}$
Now volume flow rate$=A_2V_2=\frac{\pi(0.64\times 10^{-2})^2}{4}\times 0.477=1.5\times 10^{-5}\frac{m^3}{s}$
