Chapter 13 - The Chemistry of Solutes and Solutions - Problem Solving Practice 13.5 - Page 580: a

$3.0\,\mu g$

Mass of water= density$\times$volume$=1.00\,g/mL\times100.\,mL=100\,g=0.1\,kg$ $30\,ppb\,Se\,in \,water= \frac{30\,\mu g}{1\,kg\,water}$ $\implies$ mass of Se in $\mu g$=$\frac{30\,\mu g}{1\,kg\,water}\times0.1\,kg$ $=3.0\,\mu g$