How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2

The relative leaving group ability of iodide is better than that of bromide but steric crowding is more important as the neucleophile will attack the back lobe of carbon atom.
So how should I order the steric hinderance produced by t-butyl and neopentyl groups?

The order will be $\text{neopentyl bromide} > \text{t-butyl bromide}$. 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): $$\begin{array}{c|c} \textbf{Type of halide} & \textbf{Relative Rate} \\ \hline \text{Primary Halide} & 1 \\ \text{Secondary Halide} & 0.03 \\ \text{Neopentyl Halide} & 0.00001 \\ \text{Tertiary Halide} & \approx 0 \\ \end{array}$$