The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T
(kelvin) and mass m is

de-Broglie wavelength associated with a moving particle can be given as

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

At thermal equilibrium, temperature of neutron and heavy water will be same. This common temperature is given as T. 
Also, we know that, kinetic energy of a particle,  $\mathrm{KE}=\frac{p^{\underset{¯}{2}}}{2m}$

where, p = momentum of the particle
and m = mass of the particle.
Kinetic energy of the neutron at T is  $\mathrm{KE}=\frac{3}{2}kT$

$\therefore$  de-Broglie wavelength of the neutron,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m\left(\mathrm{KE}\right)}}$

$=\frac{h}{\sqrt{2m\times (3/2)kT}}=\frac{h}{\sqrt{3mkT}}$