Among the given compounds I-III, the correct order of bond dissociation energy of C–H bond marked with * is:

The correct option is II > I > III.

To determine the correct order of the bond dissociation energy (BDE) of the C–H bonds marked with an asterisk (*), we must analyze the hybridization of the carbon atom involved in each marked bond:
1. Compound I: The marked hydrogen is attached to a carbon atom in the benzene ring. This carbon atom is ${\mathrm{sp}}^2$ hybridized.
2. Compound II: The marked hydrogen is attached to the terminal carbon of the ethynyl (alkyne) group. This carbon atom is $\mathrm{sp}$ hybridized.
3. Compound III: The marked hydrogen is attached to the saturated methylene carbon of the cyclopropene ring. This carbon atom is ${\mathrm{sp}}^3$ hybridized.

The bond dissociation energy of a C–H bond is directly related to the s-character of the hybridized orbital of the carbon atom. A higher percentage of s-character means that the bonding electrons are held closer to the carbon nucleus, resulting in a shorter, stronger bond that requires more energy to cleave homolytically.

The s-character for different hybridization states is as follows:

$\mathrm{sp}\text{hybridization}=50\%\text{s-character}$
${\mathrm{sp}}^2\text{hybridization}=33.3\%\text{s-character}$
${\mathrm{sp}}^3\text{hybridization}=25\%\text{s-character}$

Since the bond strength and dissociation energy increase with increasing s-character, the general order of bond dissociation energy for C–H bonds is:
$C\left(\mathrm{sp}\right)-H>C\left({\mathrm{sp}}^2\right)-H>C\left({\mathrm{sp}}^3\right)-H$

By comparing the marked C–H bonds in the given compounds:
- Compound II involves a $C\left(\mathrm{sp}\right)-H$ bond (highest s-character, strongest bond).
- Compound I involves a $C\left({\mathrm{sp}}^2\right)-H$ bond.
- Compound III involves a $C\left({\mathrm{sp}}^3\right)-H$ bond (lowest s-character, weakest bond).

Thus, the correct order of bond dissociation energy is II > I > III.