How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2
The order will be $\text{neopentyl bromide} &gt; \text{t-butyl bromide}$. This is just an addition to @user377386's answer. From Solomons and Fryhle's Organic Chemistry, the relative rates of reaction (SN2) of different kinds of halide are (for the same halogen):
$$ \begin{array}{c|c}
 \textbf{Type of halide} &amp; \textbf{Relative Rate} \\ \hline
 \text{Primary Halide} &amp; 1 \\
 \text{Secondary Halide} &amp; 0.03 \\
 \text{Neopentyl Halide} &amp; 0.00001 \\
 \text{Tertiary Halide} &amp; \approx 0 \\
\end{array} $$

You can easily see that neopentyl halides and tertiary halides are almost unreactive towards SN2 reaction, so it makes little sense to compare them. Neither of them is going to react by SN2 ever, they will instead react through SN1 mechanism. The slightly increased leaving group ability of iodine does not matter much. If you insist on comparing though, neopentyl halide will be better at SN2.