Finding molar mass of uranium fluoride using Graham’s law of diffusion Problem

A sample of uranium fluoride is found to effuses at the rate of $\pu{17.7 mg/h}.$ Under comparable conditions, gaseous iodine effuses at the rate of $\pu{15.0 mg/h}.$ What is the molar mass of the uranium fluoride?

Answer

 $\pu{354 g mol^-1}$

My approach
Perhaps I’m still having issues with significant figures, but I am getting the wrong answer. I know to use Graham’s law of diffusion, which leaves me with:
$$\frac{\pu{17.7 mg h^-1}}{\pu{15.0 mg h^-1}} = \sqrt\frac{\pu{253.8 g mol^-1}}{x}$$
I get $x = \pu{181.70 g mol^-1}.$ I know that the molecular mass of uranium fluoride should be less than iodine, so the answer makes no sense to me.
What am I doing wrong?
Graham's law of diffusion deals with the molar effusion rate rather than mass effusion rate.
This can be confirmed by the description of Graham's Law on Wikipedia.
An excerpt from the aforementioned page,

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$
where:
$r_1$ is the rate of effusion for the first gas (volume or amount per unit time)
$r_2$ is the rate of effusion for the second gas
$M_1$ is the molar mass of gas 1
$M_2$ is the molar mass of gas 2

Now the mathematical part,
$$\begin{align}
r_{\ce{UF6}} &amp;= \pu{17.7 mg/h} = \frac{\pu{17.7 mmol/h}}{M_{\ce{UF6}}} \\
r_{\ce{I2}} &amp;= \pu{15 mg/h} = \frac{\pu{15 mmol/h}}{M_{\ce{I2}}}
\end{align}$$
By Graham's Law,
$$\begin{align}
\frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{15} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\
\frac{17.7}{15} &amp;= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\
M_{\ce{UF6}} &amp;= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\
  &amp;= \pu{353.39 g mol^-1}
\end{align}$$