Answer
$A_2=0.25\,\mathrm{m^2}$
Work Step by Step
It can be seen from the graph that the pressure difference becomes zero when $A_1^{-2}=16\implies A_1={1\over\sqrt{16}}=0.25\,\mathrm{m^2}$.
And this can happen only if the areas of both pipe sections are equal.
Hence $A_2=A_1=0.25\,\mathrm m^2$.
