Condition for geometrical isomerism around a double bond
Geometrical isomerism ($cis$-$trans$ / $E$-$Z$) is possible only when each carbon atom of the $C=C$ bond carries two different substituents.

If on either carbon of the double bond, the two attached groups are identical, then geometrical isomerism is not possible.

Check each marked double bond
Since the structure has four marked double bonds $I, II, III,$ and $IV$, we inspect each one using the above rule.

Site $I$
At double bond $I$, one of the double-bonded carbons has two identical substituents, so the required condition for geometrical isomerism fails.
Therefore, geometrical isomerism is not possible at site $I$.

Sites $II, III,$ and $IV$
For each of these double bonds, both carbons of the $C=C$ have two different groups attached. Hence they satisfy the condition for geometrical isomerism.
Therefore, geometrical isomerism is possible at $II, III,$ and $IV$.

Conclusion
The only site where geometrical isomerism is not possible is: $I$

Match with options
Option B: $I$ is correct.