Answer
$R=6.12\,\mathrm{m^3/s}$
Work Step by Step
Using Bernoulli's equation: \begin{align*}
p_1+{1\over 2}\rho v_1^2&=p_2+{1\over 2}\rho v_2^2\\
p_2-p_1&={1\over 2}\rho v_1^2-{1\over 2}\rho v_2^2\\
p_2-p_1&=\left({1\over 2}\rho v_2^2A_2^2\right)A_1^2-{1\over 2}\rho v_2^2
\end{align*}
In the last step above we have used the continuity equation $A_1v_1=A_2v_2$.
Observe that the equation obtained is the equation of a straight line in the form $y=mx+c$, with the y-intercept $c=-{1\over 2}\rho v_2^2$. From the graph, we can clearly see that $c=-300\,\mathrm{kN/m^2}$.
Hence, ${1\over 2}\rho v_2^2=300000\implies v_2^2={600000\over 1000}=600 \implies v_2=\sqrt{600}=24.49\, \mathrm{m/s}$.
We can find the rate of flow using the equation of continuity: \begin{align*}
R&=A_2v_2\\
&=0.25\times 24.49\\
&=6.12\,\mathrm{m^3/s}
\end{align*}
