Key idea: free-radical bromination in presence of light
Bromination with $\mathrm{Br_2/h\nu}$ proceeds by a free-radical substitution mechanism. In such reactions, the hydrogen that is removed most easily is the one whose abstraction forms the most stable radical.

Rule for selectivity of bromination
Bromine is highly selective. So, the preferred hydrogen is the one that gives a radical stabilized by:

• resonance,
• benzylic stabilization,
• allylic stabilization,
• or greater alkyl substitution.

Application to compound (E)
In compound $(E)$, the hydrogen labeled as $\gamma$-hydrogen is the one whose removal leads to the most stabilized radical intermediate during bromination. Hence, that hydrogen is replaced most easily.

Evaluation of options

1. A: $\beta$-hydrogen — not the most favorable.
2. B: $\alpha$-hydrogen — not the most favorable.
3. C: $\gamma$-hydrogen — most easily replaceable.
4. D: $\delta$-hydrogen — not the most favorable.

Final answer
Therefore, the hydrogen easily replaceable during bromination in presence of light is: $\boxed{\gamma\text{-hydrogen}}$ So the correct option is C.