Answer
(a) $V = 240 \times 10^3 \,\text{V}$
(b) $V = -30 \times 10^3 \,\text{V}$
(c) $V = -90 \times 10^3 \,\text{V}$
Work Step by Step
(a) $R$ is the radius of the sphere and $r$ is the distance where the potential is measured.
When ($r R$), the potential is
\begin{equation}
V = \dfrac{1}{4 \pi \epsilon_{\circ}} \dfrac{q}{r}
\end{equation}
$r$ = $2.5 \text{cm}$, the potenatil $V$ is
\begin{align*}
V &= \dfrac{1}{4 \pi \epsilon_{\circ}} \left(\dfrac{q_{1}}{R_{1}} + \dfrac{q_{2}}{R_{2}} \right) \\
&= \left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{3.00 \times 10^{-6} \mathrm{C}}{0.05 \mathrm{m}}+\frac{-5.00 \times 10^{-6} \mathrm{C}}{0.15 \mathrm{m}}\right)\\
&=\boxed{240 \times 10^3 \,\text{V}}
\end{align*}
(b) $r$ = 10 cm the potenatil $V$ is
\begin{align*}
V&= \dfrac{1}{4 \pi \epsilon_{\circ}} \left(\dfrac{q_{1}}{r} + \dfrac{q_{2}}{R_{2}} \right) \\
&= \left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{3.00 \times 10^{-6} \mathrm{C}}{0.10 \mathrm{m}}+\frac{-5.00 \times 10^{-9} \mathrm{C}}{0.15 \mathrm{m}}\right)\\
&=\boxed{-30 \times 10^3 \,\text{V}}
\end{align*}
(c) $r$ = 20 cm the potenatil $V$ is
\begin{align*}
V &= \dfrac{1}{4 \pi \epsilon_{\circ}} \left(\dfrac{q_{1}}{r} + \dfrac{q_{2}}{r} \right) \\
&= (9.0 \times 10^{9} \mathrm{~N\cdot m^{2}/C^{2}}) \left(\dfrac{ 3.00 \times 10^{-9} \,\text{C}}{0.20 \,\text{m}} + \dfrac{- 5.00 \times 10^{-9} \,\text{C}}{0.20 \,\text{m}} \right) \\
&=\boxed{-90 \times 10^3 \,\text{V}}
\end{align*}
