Answer
$-\sqrt {\frac{3-2\sqrt 2}{6}}$
Work Step by Step
As $B$ is in QIII, $\cos B$ is negative.
$\cos B=-\sqrt {1-\sin^{2}B}$
$=-\sqrt {1-(-\frac{1}{3})^{2}}=-\sqrt {\frac{8}{9}}$
$=-\frac{2\sqrt 2}{3}$
$\cos \frac{B}{2}=\pm\sqrt {\frac{1+\cos B}{2}}$
$=\pm \sqrt {\frac{1-\frac{2\sqrt 2}{3}}{2}}=\pm\sqrt {\frac{3-2\sqrt 2}{6}}$
We need to know where $\frac{A}{2}$ terminates to determine the sign of $\cos \frac{B}{2}$.
Given that $B$ is in QIII:
Or $180^{\circ}\lt B\lt270^{\circ}$
Then $\frac{180^{\circ}}{2}\lt\frac{B}{2}\lt\frac{270^{\circ}}{2}$
That is, $90^{\circ}\lt\frac{B}{2}\lt135^{\circ}$
Therefore,
$\cos \frac{B}{2}=-\sqrt {\frac{3-2\sqrt 2}{6}}$