Answer
(a) $\left(\frac{Q_{B}}{Q_{A}}\right)=\frac{1}{3}$
(b) $\frac{E_{B}}{E_{A}} = 3$
Work Step by Step
(a) The potential in both spheres is the same and the $R_A = 3R_B$
\begin{gather}
V_{A}=V_{B}\\
k \frac{q_{A}}{R_{A}}=k \frac{q_{B}}{R_{B}}\\
\frac{q_{B}}{ q_{A}}=\frac{R_{B}}{R_{A}}\\
\left(\frac{q_{B}}{q_{A}}\right)=\frac{1}{3}
\end{gather}
(b) The electric field of the sphere is given by
$$E=k\frac{|q|}{R^{2}}$$
For both sphere, the ratio of the electric field between them will be
\begin{align}
\frac{E_{B}}{E_{A}}&=\left(\frac{q_B}{R_{B}^{2}}\right)\left(\frac{ R_{A}^{2}}{q_{A}}\right)\\
&= \left(\frac{q_B}{q_A}\right)\left(\frac{ R_{A}^{2}}{R_{B}^{2}}\right)\\
&= \frac{1}{3} \times 3^2 = 3
\end{align}
