Let L and A be length and area of cross-section of each block.

$$\therefore$$ Thermal resistance of block 1 is

$$\therefore$$ $${R_1} = {L \over {\kappa A}}$$

Thermal resistance of block 2 is

$${R_2} = {L \over {2\kappa A}}$$

In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is

$${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$$ ...... (i)

$$\therefore$$ Rate of heat flow in configuration I is

$$\therefore$$ $${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$$ ..... (ii)

In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is

$${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$$ ...... (iii)

$${R_P} = {1 \over 3}{L \over {\kappa A}}$$

$$\therefore$$ Rate of same amount of heat flow in configuration II is

$$\therefore$$ $${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$$ ...... (iv)

Divide (ii) by (iv), we get

$${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} \times {{2\kappa A} \over {3L}} = {2 \over 9}$$ (Using (i) and (iii))

$$t' = {2 \over 9}t = {2 \over 9} \times 9s = 2s$$