Solution:

To find the equivalent thermal conductivity (\(K_{\text{eq}}\)) of the compound slab with two materials in series, we use the formula for thermal resistances in series.

Formula:
The total thermal resistance \(R_{\text{total}}\) for two materials in series is:

\[
R_{\text{total}} = R_1 + R_2 = \frac{L_1}{K_1 A} + \frac{L_2}{K_2 A}
\]

Where \(L_1 = L_2\) (equal thickness) and \(A\) is the cross-sectional area (same for both). The equivalent conductivity \(K_{\text{eq}}\) is given by:

\[
R_{\text{total}} = \frac{2L}{K_{\text{eq}} A}
\]

Given:
- Thickness of each layer, \(L_1 = L_2 = L/2\)
- Thermal conductivities \(K_1 = K\) and \(K_2 = 2K\)

Substitute into the resistance formula:

\[
R_{\text{total}} = \frac{L/2}{K A} + \frac{L/2}{2K A} = \frac{L}{2KA} + \frac{L}{4KA} = \frac{3L}{4KA}
\]

Now, equate this to the total resistance for the equivalent conductivity:

\[
\frac{2L}{K_{\text{eq}} A} = \frac{3L}{4KA}
\]

Solving for \(K_{\text{eq}}\):

\[
K_{\text{eq}} = \frac{4K}{3}
\]

Thus, the equivalent thermal conductivity of the slab is:

\[
K_{\text{eq}} = \frac{4K}{3}
\]