Two metal spheres, one of radius R and the other of radius 2 R...

The surface charge density of a closed surface area having charge $\mathrm{Q}$ is given by $$\sigma=\frac{\text{ Charge }}{\text{ Area }}=\frac{\mathrm{Q}}{\mathrm{A}} \text{ or } \mathrm{Q}=\sigma \mathrm{A}$$ Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by
$$\begin{gathered} \mathrm{Q}_{\mathrm{P}}=\sigma \times 4 \pi \mathrm{R}^2=4 \pi \sigma \mathrm{R}^2 \\ \text{ and } \mathrm{Q}_{\mathrm{Q}}=\sigma \times 4 \pi(2 \mathrm{R})^2=16 \pi \sigma \mathrm{R}^2 \end{gathered}$$ when they are brought in contact with each other, the total charge will be \begin{aligned} \mathrm{Q}_{\mathrm{t}} & =\mathrm{Q}_{\mathrm{P}}+\mathrm{Q}_{\mathrm{Q}} \\ = & 4 \pi \sigma \mathrm{R}^2+16 \pi \sigma \mathrm{R}^2 \text{ [From Eq. (i) and (ii)] } \\ & =20 \pi \sigma \mathrm{R}^2 \end{aligned} In connection of two charged conducting bodies, the potential will become same on both, i.e. \begin{aligned} \frac{\mathrm{Q}_{\mathrm{P}}}{4 \pi \in_0 \mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{4 \pi \epsilon_0 2 \mathrm{R}} \\ \Rightarrow \quad \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{2 \mathrm{R}} \Rightarrow \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{Q}_{\mathrm{Q}}}=\frac{1}{2} \end{aligned} So, the charges on the sphere $\mathrm{P}$ and $\mathrm{Q}$ after separation will be distributed as $$\Rightarrow \quad Q_P^{\prime}=\frac{1}{3} Q_{\mathrm{t}} \text{ and } \mathrm{Q}_{\mathrm{Q}}^{\prime}=\frac{2}{3} \mathrm{Q}_{\mathrm{t}}$$ After separation, the new surface charge densities on $\mathrm{P}$ and $\mathrm{Q}$ will be \begin{aligned} \sigma_{\mathrm{P}}=\frac{\mathrm{Q}_{\mathrm{P}}^{\prime}}{\text{ Area }} & =\frac{1}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text{ Area }}=\frac{1}{3} \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi \mathrm{R}^2}=\frac{5}{3} \sigma \\ \text{ and } \sigma_{\mathrm{Q}}=\frac{\mathrm{Q}_{\mathrm{Q}}^{\prime}}{\text{ Area }} & =\frac{2}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text{ Area }}=\frac{2}{3} \times \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi(2 \mathrm{R})^2} \\ & =\frac{2}{3} \times \frac{5}{4} \sigma=\frac{5}{6} \sigma \end{aligned}