Notes

The smallest positive unique number that we must add to result the directory doubling in size is 6.
6 has a binary of 0110. When it is initially added to the existing hash indexes, it would fit in Bucket A2 as the rightmost 2 bits are 10. Since A2 is filled, it would cause an overflow. The local depth of A2 is equivalent to the global depth 2. This would cause the global depth to increase to 3.
A new bucket A3(say) is added in response.
Upon increasing, we would then be considering the rightmost 3 bits. The numbers from the bucket A2 are:

14 - 1110
30 - 11110
6 – 110

We can see that the rightmost 3 bits of the above number are similar i.e., each end with the bit combination of 110 and hence they would be assigned to the same newly created bucket A3.
This would cause an overflow again, and since the local depth 3 is equal to global depth 3, it would cause the global depth to increase by 1 to a new depth of 4.
The overflow would create a new bucket A4(say).
We now consider the rightmost 4 bits. Representing the numbers in their binary again:

14 – 1110
30 – 11110
6 – 0110

This would cause 6 to be placed in bucket A3 and 14 and 30 to be placed in the newly created bucket A4.
After inserting 6, we can see that the global depth/directory size has doubled from 2 to a new value of 4.