Step-by-step Solution:

1. Identify Given Information:

   • Lens type: Plano-convex lens
   • Image is real, so it's formed on the opposite side of the lens from the object.
   • Image distance, v = +8 m (positive by sign convention for a real image).
   • Image size is one-third the object size. Since the image is real, it must be inverted. Therefore, the magnification m = -1/3.
   • Wavelength relation: $\lambda_{\text{lens}} = \frac{2}{3} \lambda_{\text{free space}}$.
   • We need to find the radius of the curved surface, R.

2. Calculate the Object Distance (u): The magnification formula for a lens is $m = \frac{v}{u}$. Substituting the given values: $-\frac{1}{3} = \frac{+8 \text{ m}}{u}$ $u = -3 \times 8 \text{ m} = -24 \text{ m}$ The negative sign indicates that the object is placed 24 m in front of the lens, which is expected for a real object forming a real image with a convex lens.

3. Calculate the Focal Length (f): Using the thin lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Substituting the values of v and u: $\frac{1}{f} = \frac{1}{8} - \frac{1}{-24}$ $\frac{1}{f} = \frac{1}{8} + \frac{1}{24}$ $\frac{1}{f} = \frac{3 + 1}{24} = \frac{4}{24} = \frac{1}{6}$ $f = 6 \text{ m}$ The positive focal length confirms that it is a converging (convex) lens.

4. Calculate the Refractive Index (n) of the Lens: The refractive index n of a medium is the ratio of the wavelength of light in free space ($\lambda_{\text{free space}}$) to the wavelength of light in the medium ($\lambda_{\text{lens}}$). $n = \frac{\lambda_{\text{free space}}}{\lambda_{\text{lens}}}$ Given that $\lambda_{\text{lens}} = \frac{2}{3} \lambda_{\text{free space}}$: $n = \frac{\lambda_{\text{free space}}}{\frac{2}{3} \lambda_{\text{free space}}} = \frac{1}{2/3} = \frac{3}{2} = 1.5$

5. Use the Lens Maker's Formula to find R: The Lens Maker's formula is: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ For a plano-convex lens, one surface is planar and the other is spherical. Let the radius of the planar surface be $R_1$ and the curved surface be $R_2$. For a planar surface, the radius of curvature is infinite, so $R_1 = \infty$. Let the radius of curvature of the curved surface be R. So, $R_2 = -R$ (by sign convention, assuming light enters from the planar side) or $R_1 = R$ and $R_2 = \infty$ (assuming light enters from the curved side). In both cases, the formula simplifies. Let's assume light enters from the curved side: $R_1 = R$ and $R_2 = \infty$. $\frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n-1}{R}$ Now, substitute the values of f and n we found: $\frac{1}{6} = \frac{1.5 - 1}{R}$ $\frac{1}{6} = \frac{0.5}{R} = \frac{1/2}{R}$ $R = 6 \times \frac{1}{2} = 3 \text{ m}$

   The radius of the curved surface of the lens is 3 m.

6. Conclusion: The calculated radius of the curved surface is 3 m, which corresponds to option C.