The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.

If $K_1 = 1.25 K_2$, the value of $K_1$ is:

Solution:

Let the area of the plates be $A$ and the separation be $d$. The capacitance of the parallel plate capacitor with vacuum/air between the plates is $C_0 = \frac{\epsilon_0 A}{d}$.

When two dielectric slabs are inserted, the total thickness of the dielectrics is $t_1 = \frac{3}{8}d$ and $t_2 = \frac{d}{2}$.

The total thickness of the dielectric material is $t_{total} = t_1 + t_2 = \frac{3}{8}d + \frac{d}{2} = \frac{3d + 4d}{8} = \frac{7}{8}d$.

The remaining air gap is $t_{air} = d - t_{total} = d - \frac{7}{8}d = \frac{1}{8}d$.

The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs and an air gap is given by:

$C = \frac{\epsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + \frac{t_{air}}{1}}$

Substitute the given thicknesses:

$C = \frac{\epsilon_0 A}{\frac{3d/8}{K_1} + \frac{d/2}{K_2} + \frac{d/8}{1}}$

$C = \frac{\epsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{1}{2K_2} + \frac{1}{8} \right)}$

We know $C_0 = \frac{\epsilon_0 A}{d}$, so we can write:

$C = \frac{C_0}{\left( \frac{3}{8K_1} + \frac{4}{8K_2} + \frac{1}{8} \right)}$

According to the problem, the new capacitance $C$ is two times larger than $C_0$, so $C = 2C_0$.

$2C_0 = \frac{C_0}{\left( \frac{3}{8K_1} + \frac{4}{8K_2} + \frac{1}{8} \right)}$

Dividing both sides by $C_0$:

$2 = \frac{1}{\left( \frac{3}{8K_1} + \frac{4}{8K_2} + \frac{1}{8} \right)}$

This implies:

$\frac{3}{8K_1} + \frac{4}{8K_2} + \frac{1}{8} = \frac{1}{2}$

Multiply the entire equation by 8 to clear the denominators:

$\frac{3}{K_1} + \frac{4}{K_2} + 1 = 4$

$\frac{3}{K_1} + \frac{4}{K_2} = 3$

We are given the relation $K_1 = 1.25 K_2$. This can be written as $K_1 = \frac{5}{4} K_2$, or $K_2 = \frac{4}{5} K_1$.

Substitute $K_2$ in the equation:

$\frac{3}{K_1} + \frac{4}{\frac{4}{5} K_1} = 3$

$\frac{3}{K_1} + \frac{4 \times 5}{4 K_1} = 3$

$\frac{3}{K_1} + \frac{5}{K_1} = 3$

$\frac{8}{K_1} = 3$

$K_1 = \frac{8}{3}$

$K_1 \approx 2.666...$

Rounding to two decimal places, $K_1 = 2.67$. The closest option is 2.66.

The final answer is $\boxed{\text{2.66}}$