Correct Answer - Option 2 : 0.3 W/mK

__Concept:__

According to Fourier’s Law:

\(Q = - kA\frac{{dT}}{{dx}}\)

where k = Thermal conductivity, A = cross-section area, \(\frac {dT}{dx}\) = Temperature gradient

__Calculation:__

__Given:__

dx = 10 cm = 0.1 m, Q = 30 W/m2, dT = 10°C

\(Q = -kA\frac{{dT}}{{dx}}\)

\( \frac{Q}{A} =- k\frac{{dT}}{{dx}} \)

\( \Rightarrow 30 = k \times \frac{{10}}{{0.1}}\)

**∴ k = 0.3 W/mK**