m_n = mass of the neutron = m
m_d = 2m_n = mass of deuteron = 2m 
using conservation of momentum,
        mu_n = mV_n + 2mV_d 
⇒    u_n = V_n + 2V_d         ... (i)
using conservation of energy,

$\frac{1}{2}{\mathrm{mu}}_{\mathrm{n}}^2=\frac{1}{2}{\mathrm{mV}}_{\mathrm{n}}^2+\frac{1}{2}\left(2\mathrm{m}\right){\mathrm{V}}_{\mathrm{d}}^2$  

$\Rightarrow {\mathrm{u}}_{\mathrm{n}}^2={\mathrm{V}}_{\mathrm{n}}^2+2{\mathrm{V}}_{\mathrm{d}}^2$       .. (ii)

using (i),

${\mathrm{u}}_{\mathrm{n}}^2={\left({\mathrm{u}}_{\mathrm{n}}-2{\mathrm{V}}_{\mathrm{d}}\right)}^2+2{\mathrm{V}}_{\mathrm{d}}^2$

$\therefore {\mathrm{u}}_{\mathrm{n}}^2={\mathrm{u}}_{\mathrm{n}}^2-2{\mathrm{u}}_{\mathrm{n}}{\mathrm{V}}_{\mathrm{d}}+4{\mathrm{V}}_{\mathrm{d}}^2+2{\mathrm{V}}_{\mathrm{d}}^2$

⇒    3V_d = 2u_n 
    Fraction of energy acquired by deuteron after collision

$=\frac{1}{2}\frac{\left(2\mathrm{m}\right){\mathrm{V}}_{\mathrm{d}}^2}{\frac{1}{2}{\mathrm{mun}}_{\mathrm{n}}^2}=\frac{2{\mathrm{V}}_{\mathrm{d}}^2}{{\mathrm{u}}_{\mathrm{n}}^2}=2{\left(\frac{2}{3}\right)}^2=\frac{8}{9}$

i.e. $\frac{1}{9^{\text{th }}}$ of the incident energy is lost during a collision.
If  'n' collision slows down neutron energy from 531441 eV to 1 eV.
Then,   ${\left(\frac{1}{9}\right)}^{\mathrm{n}}=\frac{1}{531441}$
⇒    n = 6