Step-by-step Derivation:

1. Understand the Problem: We are given a solid sphere of radius R with a non-uniform volume charge density $\rho = K{r^a}$. We need to find the value of the constant 'a' given that the electric field at $r = R/2$ is 1/8th of the electric field at $r = R$.

2. Find the Charge Enclosed: To find the electric field inside the sphere at a distance 'r' from the center ($r \le R$), we first need to calculate the total charge $Q_{enc}$ enclosed within a spherical volume of radius 'r'. We can do this by integrating the charge density over the volume. Consider a thin spherical shell of radius $x$ and thickness $dx$. Its volume is $dV = 4\pi x^2 dx$. The charge $dq$ in this shell is: $dq = \rho(x) \cdot dV = (Kx^a)(4\pi x^2 dx) = 4\pi K x^{a+2} dx$ The total charge enclosed within a radius 'r' is the integral from 0 to r: $Q_{enc}(r) = \int_0^r dq = \int_0^r 4\pi K x^{a+2} dx$ $Q_{enc}(r) = 4\pi K \left[ \frac{x^{a+3}}{a+3} \right]_0^r$ $Q_{enc}(r) = \frac{4\pi K}{a+3} r^{a+3}$

3. Apply Gauss's Law: Now we use Gauss's Law to find the electric field $E(r)$ at a distance 'r' from the center. For a spherical Gaussian surface of radius 'r' ($r \le R$): $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$ Due to spherical symmetry, the electric field is radial and has a constant magnitude on the Gaussian surface. So, $\oint \vec{E} \cdot d\vec{A} = E(r) \cdot (4\pi r^2)$. $E(r) \cdot 4\pi r^2 = \frac{1}{\epsilon_0} \left( \frac{4\pi K}{a+3} r^{a+3} \right)$ Solving for $E(r)$: $E(r) = \frac{1}{4\pi r^2 \epsilon_0} \frac{4\pi K}{a+3} r^{a+3}$ $E(r) = \frac{K}{\epsilon_0(a+3)} r^{a+1}$ This expression shows that the electric field inside the sphere is proportional to $r^{a+1}$.

4. Use the Given Condition: The problem states that the electric field at $r = R/2$ is 1/8 times the field at $r = R$. Mathematically: $E(r=R/2) = \frac{1}{8} E(r=R)$ Using our derived expression for $E(r)$: $E(R/2) = \frac{K}{\epsilon_0(a+3)} \left(\frac{R}{2}\right)^{a+1}$ $E(R) = \frac{K}{\epsilon_0(a+3)} R^{a+1}$ Substituting these into the condition: $\frac{K}{\epsilon_0(a+3)} \left(\frac{R}{2}\right)^{a+1} = \frac{1}{8} \left( \frac{K}{\epsilon_0(a+3)} R^{a+1} \right)$

5. Solve for 'a': We can cancel the common terms $\frac{K}{\epsilon_0(a+3)}$ from both sides: $\left(\frac{R}{2}\right)^{a+1} = \frac{1}{8} R^{a+1}$ $\frac{R^{a+1}}{2^{a+1}} = \frac{1}{8} R^{a+1}$ Canceling $R^{a+1}$ from both sides (since $R \neq 0$): $\frac{1}{2^{a+1}} = \frac{1}{8}$ This implies: $2^{a+1} = 8$ Since $8 = 2^3$, we can write: $2^{a+1} = 2^3$ Equating the exponents: $a+1 = 3$ $a = 2$

Thus, the value of 'a' is 2.