From the previous question, $2{\int }_0^{\pi /2}f(x)g(x)dx={\int }_0^{\pi /2}g(x)dx$. So it suffices to evaluate the integral of $g$.

$g(x)=\sqrt{\frac{\pi x}{2}-x^2}=\sqrt{{\left(\frac{\pi }{4}\right)}^2-{\left(x-\frac{\pi }{4}\right)}^2}.$

This describes a semicircle of radius $\frac{\pi }{4}$ centred at $x=\frac{\pi }{4}$ on the x-axis. The integral ${\int }_0^{\pi /2}g(x)dx$ equals the area of a semicircular disc of radius $\frac{\pi }{4}$:

${\int }_0^{\pi /2}g(x)dx=\frac{1}{2}\pi {\left(\frac{\pi }{4}\right)}^2=\frac{{\pi }^3}{32}.$

Therefore ${\int }_0^{\pi /2}f(x)g(x)dx=\frac{1}{2}\cdot \frac{{\pi }^3}{32}=\frac{{\pi }^3}{64}.$

Multiplying by $\frac{16}{{\pi }^3}$:

$\frac{16}{{\pi }^3}\cdot \frac{{\pi }^3}{64}=\frac{1}{4}=\mathrm{0.25.}$