1. Find the refractive index of the medium

Given that the wavelength inside the medium is $\dfrac{2}{3}$ times the wavelength in air:

$\lambda_{\text{medium}}=\frac{2}{3}\lambda_{\text{air}}$

Since frequency remains unchanged,

$n=\frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}}=\frac{1}{2/3}=\frac{3}{2}$

So, the refractive index of the refracting medium is

$n_2=\frac{3}{2}, \qquad n_1=1$

2. Use the given image distance relation

The image is formed 10 m behind the surface, and it is real. So,

$v=+10\,\text{m}$

Also, the image distance is $\dfrac{2}{3}$ of the object distance from the surface:

$v=\frac{2}{3}u_{\text{distance}}$

Hence,

$10=\frac{2}{3}u_{\text{distance}} \Rightarrow u_{\text{distance}}=15\,\text{m}$

Using Cartesian sign convention, object is on the left of the surface:

$u=-15\,\text{m}$

3. Apply refraction at a spherical surface formula

For refraction at a spherical surface:

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Substitute values:

$\frac{3/2}{10}-\frac{1}{-15}=\frac{(3/2)-1}{R}$

$\frac{3}{20}+\frac{1}{15}=\frac{1}{2R}$

Take LCM $60$:

$\frac{9+4}{60}=\frac{1}{2R}$

$\frac{13}{60}=\frac{1}{2R}$

So,

$2R=\frac{60}{13}$

$R=\frac{30}{13}\,\text{m}$

4. Compare with the given form

Radius is given as

$R=\frac{x}{13}\,\text{m}$

Thus,

$\frac{x}{13}=\frac{30}{13} \Rightarrow x=30$

5. Final answer

$\boxed{30}$