Finding molar mass of uranium fluoride using Graham’s law of diffusion
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,
$$\begin{align}
\frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{15} &amp;= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\
\frac{17.7}{15} &amp;= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\
M_{\ce{UF6}} &amp;= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\
  &amp;= \pu{353.39 g mol^-1}
\end{align}$$