Three composite rectangular slabs A, B, and C of equal thickness...

The thermal resistance of a slab is given by $R = \dfrac{L}{kA}$.

For the three slabs A, B, and C having equal thickness $L$ and equal cross-sectional area $A$, the thermal resistances are:
$R_A = \dfrac{L}{3KA}$
$R_B = \dfrac{L}{2KA}$
$R_C = \dfrac{L}{KA}$

Since the slabs are connected in series, the rate of heat flow $H$ is the same through all of them. The temperature difference across any slab is proportional to its thermal resistance ($\Delta T = HR$).

The ratio of temperature differences across the slabs is:
$\Delta T_A : \Delta T_B : \Delta T_C = R_A : R_B : R_C = \dfrac{1}{3} : \dfrac{1}{2} : 1 = 2 : 3 : 6$

The total temperature difference across the composite slab is:
$\Delta T_{total} = 66^\circ\text{C} - 0^\circ\text{C} = 66^\circ\text{C}$

The temperature difference across the middle slab B is:
$\Delta T_B = \dfrac{3}{2 + 3 + 6} \times \Delta T_{total} = \dfrac{3}{11} \times 66^\circ\text{C} = 18^\circ\text{C}$

Answer: $18^\circ\text{C}$