working with ratios and RMS to find variables
As I said solve using variables first and simplify as much as you can, then plug in the numbers. From the Maxwell-Boltzmann Distribution:

$v_p = \sqrt{\frac{2RT}{M}}$

$v_{rms} = \sqrt{\frac{3}{2}}v_p$

where:


$v_p$ most probably velocity
$v_{rms}$ rms velocity
R universal gas constant
T temperature
M molecular mass


so let:


$v_{p,He}$ - most probably velocity for He 
$v_{p,U}$ - most probably velocity for $\ce{UF6}$ 
$v_{rms,He}$ - rms velocity for He 
$v_{rms,U}$ - rms probably velocity for $\ce{UF6}$ 


we're given:
 $v_{rms,He} = v_{rms,U}$

but $v_{rms,He} = \sqrt{\frac{3}{2}}v_{p,He}$
and $v_{rms,U} = \sqrt{\frac{3}{2}}v_{p,U}$
so:
$\sqrt{\frac{3}{2}}v_{p,He} = \sqrt{\frac{3}{2}}v_{p,U}$
which can be simplified to: 
$v_{p,He} = v_{p,U}$ 

Now substituting from the Maxwell-Boltzmann equation for $v_p$ for both He and $\ce{UF6}$ we have:

$\sqrt{\frac{2RT_{He}}{M_{He}} } =  \sqrt{\frac{2RT_U}{M_U}}$ 

simplying
$\frac{T_{He}}{M_{He}} = \frac{T_{U}}{M_{U}}$

rearranging:
$\frac{T_{U}}{T_{He}} = \frac{M_{U}}{M_{He}} = \frac{352.02}{4.003} = 87.94$

Superfluous information is that $\ce{UF6}$ decomposes about $1800^o \text{ K}$ and is a solid below about $330^o \text{ K}$. Helium liquefies about $4.2^o \text{ K}$. So this only works if Helium is in range $4.2^o \text{ K}$ to $18.4^o \text{ K}$ which would mean that the uranium hexfluoride is between $369^o \text{ K}$ to $1800^o \text{ K}$.