$\mathrm{R}=\int \mathrm{dR}=\frac{\rho}{4 \pi} \int_{\mathrm{r}_{1}}^{\mathrm{r}_{2}} \frac{\mathrm{dr}}{\mathrm{r}^{2}} \Rightarrow \mathrm{R}=\frac{\rho\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)}{4 \pi \mathrm{r}_{1} \mathrm{r}_{2}}$

Rate of melting is max when power dissipated in sphere is max. Using maximum power transfer theorem,

$\mathrm{R}=\mathrm{r} .$ of battery i.e. $\frac{\rho\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)}{4 \pi \mathrm{r}_{1} \mathrm{r}_{2}}=\frac{2}{\pi}$

$\rho  = \frac{{8{r_1}{r_2}}}{{{r_2} - {r_1}}} = \frac{{8(200)}}{{10}} = \frac{{160}}{{100}}$  (in $SI$)

Also, $\sigma=\frac{1}{\rho}=\frac{10}{16}=\frac{5}{8}$