a
(a) $\frac{Q}{t} = \frac{{KA\,\Delta \theta }}{l}$$ = \frac{{\Delta \theta }}{{(l/KA)}} = \frac{{\Delta \theta }}{R}$ ( $R =$ Thermal resistance)
==> $t \propto R$ ( $Q$ and $\Delta \theta $ are same)
==> $\frac{{{t_P}}}{{{t_S}}} = \frac{{{R_P}}}{{{R_S}}} = \frac{{R/2}}{{2R}} = \frac{1}{4}$
==>${t_P} = \frac{{{t_S}}}{4} = \frac{4}{4} = 1\min .$
(Series resistance ${R_S} = {R_1} + {R_2}$ and parallel resistance ${R_P} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$)