b
(b) $\frac{{dT}}{{dt}} = \frac{{\sigma \,A}}{{mcJ}}\,\,({T^4} - T_0^4)$ [In the given problem fall in temperature of body $dT = (200 - 100) = 100K$, temp. of surrounding $T0 = 0K$ , Initial temperature of body $T = 200K].$
$\frac{{100}}{{dt}} = \frac{{\sigma 4\pi {r^2}}}{{\frac{4}{3}\pi {r^3}\rho \,c\,J}}({200^4} - {0^4})$
==> $dt = \frac{{r\rho \,c\,J}}{{48\sigma }} \times {10^{ - 6}}s = \frac{{r\rho \,c}}{\sigma }.\frac{{4.2}}{{48}} \times {10^{ - 6}}$
$ = \frac{7}{{80}}\frac{{r\rho \,c}}{\sigma }\mu \,s\tilde --\frac{7}{{72}}\frac{{r\rho \,c}}{\sigma }\mu \,s$ [As $J = 4.2$ ]