A pump can fill a tank with water in $2\text{hours}$. Because of a leak developed in the tank, it takes now $2.5\text{hours}$ to fill the tank. The leak will empty the full tank with pump not running, in:

Let the filling rate of the pump be $A$ and the emptying rate of the leak be $B$.
The pump fills the tank in $2\text{hours}$, so its work rate is:

$$A=\frac{1}{2}\text{tank/hour}$$

With the leak active, the net filling rate becomes:

$$A-B=\frac{1}{2.5}=\frac{2}{5}\text{tank/hour}$$

To find the emptying rate of the leak ($B$), substitute $A$:

$$\frac{1}{2}-B=\frac{2}{5}$$

$$B=\frac{1}{2}-\frac{2}{5}=\frac{5-4}{10}=\frac{1}{10}\text{tank/hour}$$

Thus, the leak alone will empty the full tank in:

$$\text{Time}=\frac{1}{B}=10\text{hours}$$