To solve this problem, we need to determine the time \( t \) after which tap B should be opened so that when the tank is full, the amount of water from tap A is double that from tap B.

Let's denote:
- \( T \) as the total time it takes to fill the tank.
- \( t \) as the time after which tap B is opened.

First, we calculate the rates at which taps A and B fill the tank:
- Tap A fills the tank in 23 minutes, so its rate is \( \frac{1}{23} \) of the tank per minute.
- Tap B fills the tank in 16 minutes, so its rate is \( \frac{1}{16} \) of the tank per minute.

Since tap A is opened first and tap B is opened \( t \) minutes later, the total time \( T \) to fill the tank can be expressed as:
\[ T = t + (T - t) \]

During the time \( t \), only tap A is running. The amount of water from tap A in this period is:
\[ \text{Water from tap A in } t \text{ minutes} = t \times \frac{1}{23} \]

After \( t \) minutes, both taps A and B are running together for the remaining \( T - t \) minutes. The amount of water from tap A in this period is:
\[ \text{Water from tap A in } (T - t) \text{ minutes} = (T - t) \times \frac{1}{23} \]

The total amount of water from tap A when the tank is full is:
\[ \text{Total water from tap A} = t \times \frac{1}{23} + (T - t) \times \frac{1}{23} = T \times \frac{1}{23} \]

The amount of water from tap B in the \( T - t \) minutes it is running is:
\[ \text{Water from tap B in } (T - t) \text{ minutes} = (T - t) \times \frac{1}{16} \]

According to the problem, the amount of water from tap A should be double that from tap B when the tank is full:
\[ T \times \frac{1}{23} = 2 \times (T - t) \times \frac{1}{16} \]

Now, we solve this equation for \( t \):
\[ \frac{T}{23} = 2 \times \frac{T - t}{16} \]
\[ \frac{T}{23} = \frac{2(T - t)}{16} \]
\[ \frac{T}{23} = \frac{T - t}{8} \]

Cross-multiplying to solve for \( t \):
\[ 8T = 23(T - t) \]
\[ 8T = 23T - 23t \]
\[ 23t = 23T - 8T \]
\[ 23t = 15T \]
\[ t = \frac{15T}{23} \]

Since the tank is filled completely, we know that the combined rates of both taps must fill the tank in time \( T \):
\[ T \times \left( \frac{1}{23} + \frac{1}{16} \right) = 1 \]

Finding a common denominator for the rates:
\[ \frac{1}{23} + \frac{1}{16} = \frac{16 + 23}{368} = \frac{39}{368} \]

Thus:
\[ T \times \frac{39}{368} = 1 \]
\[ T = \frac{368}{39} \approx 9.44 \text{ minutes} \]

Now, substituting \( T \) back into the equation for \( t \):
\[ t = \frac{15T}{23} = \frac{15 \times 368}{23 \times 39} = \frac{5520}{897} \approx 6.15 \text{ minutes} \]

Therefore, tap B should be opened approximately 6.15 minutes after tap A to ensure that the amount of water from tap A is double that from tap B when the tank is full.