A is twice as efficient as B. If A can finish a piece of work in 60 days, in how many days can A and B finish the work together?

A is twice as efficient as B. This means A takes half the time B takes to complete the same work.
Given that A finishes the work in 60 days.
Let $D_A$ be the number of days A takes and $D_B$ be the number of days B takes.

$$D_A = 60 \text{ days}$$

Since A is twice as efficient as B, B will take twice as long as A.

$$D_B = 2 \times D_A$$ $$D_B = 2 \times 60 \text{ days}$$ $$D_B = 120 \text{ days}$$

The work rate of A is $\frac{1}{D_A}$ per day.
The work rate of B is $\frac{1}{D_B}$ per day.
Combined work rate = Work rate of A + Work rate of B

$$\text{Combined work rate} = \frac{1}{60} + \frac{1}{120}$$

To add these fractions, find a common denominator, which is 120.

$$\text{Combined work rate} = \frac{2}{120} + \frac{1}{120}$$ $$\text{Combined work rate} = \frac{2+1}{120}$$ $$\text{Combined work rate} = \frac{3}{120}$$ $$\text{Combined work rate} = \frac{1}{40} \text{ work per day}$$

The number of days taken to finish the work together is the reciprocal of their combined work rate.

$$\text{Days together} = \frac{1}{\text{Combined work rate}}$$ $$\text{Days together} = \frac{1}{\frac{1}{40}}$$ $$\text{Days together} = 40 \text{ days}$$

The correct option is A.

The final answer is $\boxed{\text{40 days}}$.