Let $p(x)=8 x^{4}+4 x^{3}-16 x^{2}+10 x+m.$

As $(2 x-1)$ is a factor of $p ( x )$

$\therefore \quad p\left(\frac{1}{2}\right)=0$ [By factor theorem]

$\Rightarrow \quad 8\left(\frac{1}{2}\right)^{4}+4\left(\frac{1}{2}\right)^{3}-16\left(\frac{1}{2}\right)^{2}+10\left(\frac{1}{2}\right)+m=0$

$\Rightarrow \quad 8\left(\frac{1}{16}\right)+4\left(\frac{1}{8}\right)-16\left(\frac{1}{4}\right)+5+m=0$

$\Rightarrow \quad \frac{1}{2}+\frac{1}{2}-4+5+m=0$

$\Rightarrow \quad 2+m=0 \Rightarrow m=-2$