a
Charge in layer of width $\mathrm{dr}$
$\mathrm{dq}=2 \pi \mathrm{rdr} \sigma$
Work to bring $\mathrm{dq}=\mathrm{dU}=\mathrm{Vdq}$
$\mathrm{dU}=4 \sigma r \cdot 2 \pi \mathrm{rdr} \sigma=8 \pi \sigma^{2} \mathrm{r}^{2} \mathrm{dr}$
$\mathrm{U}=8 \pi \sigma^{2} \int_{0}^{\mathrm{R}} \mathrm{r}^{2} \mathrm{dr}=\frac{8 \pi \sigma^{2}}{3} \mathrm{R}^{3}$
