Answer
$2.0\%(\text{m/v})$ KBr solution.
Work Step by Step
$C_{1}=8.0\%(\text{m/v})$
$V_{1}=50.0\,mL$
$V_{2}=200.0\,mL$
Rearranging the dilute expression $C_{1}V_{1}=C_{2}V_{2}$, we get
$C_{2}=\frac{C_{1}V_{1}}{V_{2}}=\frac{8.0\%\times50.0\,mL}{200.0\,mL}=2.0\%(\text{m/v})$
