How to compare the SN2 rate of iodocyclopropane and 1-iodobicyclo[2.2.1]heptane?
Backside attack is must, but as the cyclopropane ring is planar it's quite impossible for back side attack. Then the bridge compound is non planar, we know in bridge compound both SN1 and SN2 substitution are bad, so I just confused in these two.
Concerning the rates of SN2 reaction, the rate of reaction for cyclopropyl iodide will be relatively higher than 1-iodonorbornane (name of given bridged compound).
The cyclopropyl iodide (same as 1-iodocyclopropane) has one hydrogen atom substituted by the iodine atom. If you look into the proper stereochemistry of the compound, this $\ce{I}$ atom can be above the plane of the cyclopropane ring or below the plane. Angle between the above-planar atom and below-planar atom is approximately 115 degrees. So, the attacking nucleophile can attack from above the plane if $\ce{I}$ is below the plane, or vice-versa. This attack would be best if leaving group $\ce{I^-}$ and the nucleophile are in almost 180 degrees position. Here the repulsion due to lesser bond angle and with the hydrogen will decrease the rate, but SN2 is still possible to a low extent in the cyclopropane ring.
Now to the bridged compound. Firstly, there is very very little chance for backside attack. Secondly, for occurrence of SN2, there should be a Walden inversion of configuration. But in the briged bicyclo compound, the ring is in a very compact structure, and if it tries to undergo any inversion of configuration, a huge strain will be created in the bicyclo-compound, which is thermodynamically, very unstable. So, the rate of SN2 is almost nil for 1-iodo norborane.
