At temperature T, the average kinetic energy of any particle is $$\frac{3}{2}kT$$. The de Broglie wavelength follows the order:

We start from the statement that, for any non-relativistic particle in thermal equilibrium at temperature $$T$$, the average kinetic energy is

$$K=\frac32\,kT.$$

For a particle of rest mass $$m$$ the kinetic energy is also related to the linear momentum $$p$$ by the classical formula

$$K=\frac{p^{2}}{2m}.$$

Solving this for the momentum gives

$$p=\sqrt{2mK}.$$

The de Broglie wavelength is defined as

$$\lambda=\frac{h}{p},$$

so after substituting $$p=\sqrt{2mK}$$ we obtain

$$\lambda=\frac{h}{\sqrt{2mK}}.$$

Now we insert $$K=\dfrac32 kT$$ to express the wavelength of a thermal particle explicitly in terms of its mass:

$$\lambda=\frac{h}{\sqrt{2m\left(\dfrac32 kT\right)}} =\frac{h}{\sqrt{3mkT}}.$$

From this final form we notice the crucial dependence

$$\lambda\propto\frac1{\sqrt{m}}.$$

In simple words, for particles that possess the same thermal kinetic energy, the lighter the mass, the longer the de Broglie wavelength.

Let us now deal with each object mentioned in the problem.

1. Thermal electron

The electron mass is

$$m_e\approx9.11\times10^{-31}\,\text{kg}.$$

Its thermal de Broglie wavelength at temperature $$T$$ is therefore

$$\lambda_e=\frac{h}{\sqrt{3m_e kT}}.$$

2. Thermal neutron

The neutron mass is

$$m_n\approx1.675\times10^{-27}\,\text{kg},$$

which is about $$1836$$ times heavier than the electron. Because $$\lambda\propto1/\sqrt{m}$$ we immediately see that

$$\lambda_n=\frac{h}{\sqrt{3m_n kT}}$$

is much smaller than $$\lambda_e$$. Quantitatively, the ratio is

$$\frac{\lambda_e}{\lambda_n}=\sqrt{\frac{m_n}{m_e}}\approx\sqrt{1836}\approx43,$$

so the electron wavelength exceeds the neutron wavelength by roughly two orders of magnitude.

3. Visible photon

A photon is massless; its wavelength is fixed by its energy through the Planck relation

$$E=\frac{hc}{\lambda_{\text{ph}}}.$$

For visible light, $$\lambda_{\text{ph}}$$ lies roughly between $$400\;\text{nm}$$ and $$700\;\text{nm}$$; a central value $$\lambda_{\text{ph}}\approx500\;\text{nm}=5\times10^{-7}\text{ m}$$ is adequate for comparison.

To see the numerical contrast, let us evaluate the actual thermal wavelengths at room temperature, say $$T=300\;\text{K}$$.

Thermal electron

$$K=\frac32 kT=\frac32(1.38\times10^{-23})(300) \approx6.21\times10^{-21}\,\text{J},$$

$$p_e=\sqrt{2m_eK} =\sqrt{2(9.11\times10^{-31})(6.21\times10^{-21})} \approx1.06\times10^{-25}\,\text{kg m s}^{-1},$$

$$\lambda_e=\frac{h}{p_e} =\frac{6.626\times10^{-34}}{1.06\times10^{-25}} \approx6.2\times10^{-9}\,\text{m}=6.2\;\text{nm}.$$

Thermal neutron

$$p_n=\sqrt{2m_nK} =\sqrt{2(1.675\times10^{-27})(6.21\times10^{-21})} \approx4.56\times10^{-24}\,\text{kg m s}^{-1},$$

$$\lambda_n=\frac{h}{p_n} =\frac{6.626\times10^{-34}}{4.56\times10^{-24}} \approx1.45\times10^{-10}\,\text{m}=0.145\;\text{nm}.$$

Visible photon

We already have $$\lambda_{\text{ph}}\approx5\times10^{-7}\,\text{m}=500\;\text{nm}.$$

Collecting the three wavelengths we find

$$\lambda_{\text{photon}}\;(\sim500\text{ nm}) \; > \; \lambda_{\text{electron}}\;(\sim6\text{ nm}) \; > \; \lambda_{\text{neutron}}\;(\sim0.15\text{ nm}).$$

Thus the required descending order is

Visible photon > thermal electron > thermal neutron.

Comparing with the options supplied in the question, this sequence corresponds exactly to Option A.

Hence, the correct answer is Option A.