${p_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2$

$\therefore v_2^2 - v_1^2 = 2gh$                        $...\left( 1 \right)$

According to the equation of continuty

${A_1}{v_1} = {A_2}{v_2}$                              $...(2)$

$\frac{{{A_1}}}{{{A_2}}} = \frac{{6\,m{m^2}}}{{10\,m{m^2}}}$

$From\,equation\,\left( 2 \right),\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = \frac{6}{{16}}$

$or,\,\,{v_2} = \frac{6}{{10}}{v_1}$

Putting this value of $v_2$ in equation $(1)$

${\left( {\frac{6}{{10}}{v_1}} \right)^2} - {\left( {{v_1}} \right)^2} = 2 \times {10^3} \times 5$

                                   $\left[ \begin{array}{l}
g = 10m/{s^2} = {10^3}cm/{s^2}\\
and\,h = 5\,cm
\end{array} \right]$

$Solving\,we\,get\,{v_1} = \frac{{10}}{8}$

Therefore the rate at which water flows through the tube

$ = {A_1}{v_1} = {A_2}{v_2} = \frac{{6 \times 10}}{8} = 7.5cc/s$