Two rods of same length, radius and material transfer a given amount of heat in 't' second when they are joined as shown in fig (1). But when they are joined as shown in fig (2) then they will transfer same heat in same condition in time

Let the thermal resistance of each rod be $R$.

In figure (1), the two rods are connected in parallel. The equivalent thermal resistance is:
$R_p = \dfrac{R \times R}{R + R} = \dfrac{R}{2}$

The amount of heat transferred in time $t$ is:
$Q = \dfrac{\Delta T}{R_p} t = \dfrac{2 \Delta T}{R} t$

In figure (2), the two rods are connected in series. The equivalent thermal resistance is:
$R_s = R + R = 2R$

Let the time taken to transfer the same amount of heat $Q$ under the same temperature difference $\Delta T$ be $t'$. Then:
$Q = \dfrac{\Delta T}{R_s} t' = \dfrac{\Delta T}{2R} t'$

Equating the two expressions for $Q$:
$\dfrac{2 \Delta T}{R} t = \dfrac{\Delta T}{2R} t'$

$2t = \dfrac{t'}{2}$

$t' = 4t$

Answer: $4t$