Answer
(a) $\sigma_K = 0.62 \times 10^{-6} \mathrm{~C/m^2}$
(b) $K = 1.28$
Work Step by Step
(a) The dielectric constant equals the ration between the two electric fields in the form
$$K = \dfrac{E_o}{E} = \dfrac{3.2 \times 10^5 \,\text{V/m}}{2.5 \times 10^5 \,\text{V/m}} = 1.28 $$
The charge density $\sigma_o$ before the dielectric is given by
$$\sigma_o = \epsilon_o E_o = (8.85 \times 10^{-12} \,\text{F/m})(3.2 \times 10^5 \,\text{V/m}) = 2.8 \times 10^{-6} \mathrm{~C/m^2}$$
After the dielectric material is inserted, the charge density $\sigma_K$ is given by
\begin{align*}
\sigma_k &= \sigma_o \left( 1 - \dfrac{1}{K}\right)\\
&= 2.8 \times 10^{-6} \mathrm{~C/m^2} \left( 1 - \dfrac{1}{1.28}\right)\\
&= \boxed{0.62 \times 10^{-6} \mathrm{~C/m^2}}
\end{align*}
(b) We calculated $K $ by
$$\boxed{K = 1.28} $$
