Answer
a) $0\;\rm m/s$
b) $56.6\;\rm m/s$
c) $60\;\rm m/s$
Work Step by Step
a) The best way to solve this problem is to do as the author did in Example 18.2.
\begin{array}{|c|c|c|}
\hline
{\rm Paticle}& v_x\;{\rm (m/s)}& v_y\;{\rm (m/s)}& v_x^2\;{\rm (m/s)^2}&v_y^2\;{\rm (m/s)^2}&v^2\;{\rm (m/s)^2}&v\;{\rm (m/s)}\\
\hline
1 & 20&-30 &400 &900 &1300 & 10\sqrt{13} \\
\hline
2 &40 &70 & 1600& 4900& 6500 & 10\sqrt{65} \\
\hline
3& -80&20 &6400 &400 & 6800 & 20\sqrt{17} \\
\hline
4& 30 &0 & 900 &0 & 900 & 30 \\
\hline
5 & 40 & -40 & 1600 &1600 & 3200 &40\sqrt{2} \\
\hline
6 & -50&-20 & 2500 &400 & 2900 & 10\sqrt{29} \\
\hline
{\rm Average}& 0 & 0 & - &- & 3600 & 56.6 \\
\hline
\end{array}
where $v^2=v_x^2+v_y^2$, and hence, $v=\sqrt{v^2}$.
Also $v_{avg,x}=\dfrac{v_{x1}+....+v_{x6}}{6}$.
And hence the average velocity
$$\vec v_{avg}=\sqrt{v_{avg,x}^2+v_{avg,y}^2}=\sqrt{0+0}=\color{red}{\bf0}\;\rm m/s$$
______________________________________________________________
b) $v_{avg}$ is calculated in the table above the last down-right cell.
$$v_{avg}=\color{red}{\bf 56.6}\;\rm m/s$$
______________________________________________________________
c) Now we need to find the root-mean-square speed $v_{\rm rms}$ which is given by
$$v_{\rm rms}=\sqrt{(v^2)_{avg}}$$
Plugging from the table;
$$v_{\rm rms}=\sqrt{(3600)}=\color{red}{\bf 60}\;\rm m/s $$