As per the question, if two rods are attached end to end to each other then it takes 12 sec. Now when they are attached parallel then its length halves and area will be doubled.
Heat flow is given as:
Q = KAΔT/L
so, T ∝ L/A
T' ∝ (L/2)/(2A)
T' ∝ (L/4A)
T' /T = (L/4A)/(L/A) = 4
T' = 1/4T = 1/4 X 12 = 3 sec