Answer
(a) $E =53 \mathrm{N} / \mathrm{C} $
(b) $E =160 \mathrm{N} / \mathrm{C} $
(c) $E =480 \mathrm{N} / \mathrm{C} $
Work Step by Step
(a) The electric field $E$ is inversely proportional to the square distance $r^2$, so for two instants of the sphere we could get the electric field when $r$ = 0.2 cm by
\begin{gather}
\frac{E_{1}}{E_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}} \\
E_{2}=E_{1}\left(\frac{r_{1}}{r_{2}}\right)^2\\
E_2 = (480 \mathrm{N} / \mathrm{C})\left(\frac{0.2 \mathrm{cm}}{0.6 \mathrm{cm}}\right)^{2}\\
E =53 \mathrm{N} / \mathrm{C}
\end{gather}
(b) For the cylindrical shape, the electric field $E$ is inversely proportional to the distance $r$, so for two instants of the cylinder we could get the electric field when $r$ = 0.2 cm by
\begin{aligned}
E_{2} &=E_{1}\left(\frac{r_{1}}{r_{2}}\right) \\ &=480 \mathrm{N} / \mathrm{C}\left(\frac{0.2 \mathrm{cm}}{0.6 \mathrm{cm}}\right) \\ &=160 \mathrm{N} / \mathrm{C}
\end{aligned}
(c) For the sheet, the electric field independent in the distance, so the electric field is the same for different distance
$$E =480 \mathrm{N} / \mathrm{C} $$
