Answer
$\frac{3\sqrt {10}}{10}$
Work Step by Step
As $B$ is in QI, $\cos B$ is positive.
$\cos B=\sqrt {1-\sin^{2}B}$
$=\sqrt {1-(\frac{3}{5})^{2}}=\sqrt {\frac{16}{25}}$
$=\frac{4}{5}$
$\cos \frac{B}{2}=\pm\sqrt {\frac{1+\cos B}{2}}$
$=\pm \sqrt {\frac{1+\frac{4}{5}}{2}}=\pm \sqrt {\frac{9}{10}}=\pm\frac{3}{\sqrt {10}}=\pm\frac{3\sqrt {10}}{10}$
We need to know where $\frac{B}{2}$ terminates to determine the sign of $\cos \frac{B}{2}$.
Given that $B$ is in QI:
Or $0^{\circ}\lt B\lt90^{\circ}$
Then $\frac{0^{\circ}}{2}\lt\frac{B}{2}\lt\frac{90^{\circ}}{2}$
That is, $0^{\circ}\lt\frac{B}{2}\lt45^{\circ}$
So $\frac{B}{2}$ terminates in the first quadrant. In the first quadrant, cosine is positive. Therefore,
$\cos \frac{B}{2}=\frac{3\sqrt {10}}{ 10}$