a
(a) Heat current $H = \frac{{\Delta \theta }}{R}$ ==> $\frac{{{H_P}}}{{{H_S}}} = \frac{{{R_S}}}{{{R_P}}}$
In first case : ${R_S} = {R_1} + {R_2} = \frac{l}{{(3K)A}} + \frac{l}{{KA}} = \frac{4}{3}\frac{l}{{KA}}$
In second case :${R_P} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{\frac{l}{{(3K)A}} \times \frac{l}{{KA}}}}{{\left( {\frac{l}{{(3K)A}} + \frac{l}{{KA}}} \right)}} = \frac{l}{{4KA}}$
$\therefore \frac{{{H_P}}}{{{H_S}}} = \frac{{\frac{{4l}}{{3KA}}}}{{\frac{l}{{4KA}}}} = \frac{{16}}{3}$