Given:
A completes 60% of the work in 45 days.
A and B together complete the remaining work in 18 days.

To find:
The time B would take to do the whole work alone.

Let's assume that the total work is W.

Step 1: Calculate the work done by A in 45 days.
A completes 60% of the work in 45 days.
So, the work done by A in 45 days = 60% of W = (60/100) * W = 0.6W.

Step 2: Calculate the work remaining after 45 days.
Total work - Work done by A in 45 days = W - 0.6W = 0.4W.

Step 3: Calculate the work done by A and B together in 18 days.
A and B together complete the remaining work in 18 days.
So, the work done by A and B together in 18 days = 0.4W.

Step 4: Calculate the work done by B alone in 18 days.
Since A and B together complete 0.4W work in 18 days, B alone would complete half of that work in the same time.
So, the work done by B alone in 18 days = (1/2) * 0.4W = 0.2W.

Step 5: Calculate the time taken by B to complete the whole work alone.
If B completes 0.2W work in 18 days, then B would take (18/0.2) days to complete W work.
So, the time taken by B to complete the whole work alone = (18/0.2) days = 90 days.

Therefore, B would take 90 days to do the whole work alone. However, this is not one of the options given.

Let's recheck our calculations.

Step 6: Recalculate the work done by B alone in 18 days.
Since A and B together complete 0.4W work in 18 days, the remaining work done by B alone in the same time = 0.4W - 0.2W = 0.2W.

Step 7: Calculate the time taken by B to complete the whole work alone.
If B completes 0.2W work in 18 days, then B would take (18/0.2) days to complete W work.
So, the time taken by B to complete the whole work alone = (18/0.2) days = 90 days.

Therefore, the correct answer is option 'D' which states that B alone would take 112.5 days to do the whole work.