If the area bounded by $y^2=4ax$ and $x^2=4ay$ is $\frac{64}{3}$ square units, then the positive value of $a$ is

Point of intersection of the two given curves are $\left(\mathrm{0,0}\right)$ and $\left(4a,4a\right)$

The area bounded $=2{\int }_0^{4a}\left(x-\frac{x^2}{4a}\right)dx$
$=2{\left[\frac{x^2}{2}-\frac{x^3}{12a}\right]}_0^{4a}$
$=2\left[8a^2-\frac{16a^2}{3}\right]=\frac{16a^2}{3}$ sq. units
Given, $\frac{16a^2}{3}=\frac{64}{3}$

Hence, the positive value of $a=2$