Number of worst case comparisons in linear probing = Size of largest cluster + 1
In this case, we have two clusters of size $3$ $(abc, efg)$. Worst case can occur when we are searching for a non-existent element, that should have mapped to either the location of $a: 0$, or the location of $e: 6$.
So, number of comparisons in worst case for this Hash Table are:
Size of largest cluster + 1 $= 3 + 1 = 4$
