$\sin \theta +\cos \theta =\frac{1}{2}$

${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{1}{4}\Rightarrow \sin 2\theta =-\frac{3}{4}$

Now:

$\cos 4\theta =1-2{\sin }^{2}2\theta$

$=1-2{\left(-\frac{3}{4}\right)}^2$

$=1-2\times \frac{9}{16}=-\frac{1}{8}$

And $\sin 6\theta =3\sin 2\theta -4{\sin }^{3}2\theta$

$=\left(3-4{\sin }^{2}2\theta \right)·\sin 2\theta$

$=\left[3-4\left(\frac{9}{16}\right)\right]·\left(-\frac{3}{4}\right)$

$=\left[\frac{3}{4}\right]\times \left(-\frac{3}{4}\right)=-\frac{9}{16}$

So, $16\left[\sin 2\theta +\cos 4\theta +\sin 6\theta \right]$

$=16\left(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16}\right)=-23$