Solution:

To find the ratio of the energy stored in the capacitor to the work done by the battery, let's break it down step by step:

1. Energy Stored in the Capacitor

The energy $U$ stored in a capacitor is given by:

$$U = \frac{1}{2} C V^2$$

where $C$ is the capacitance, and $V$ is the voltage across the capacitor (equal to the EMF of the battery once fully charged).

2. Work Done by the Battery

The work done by the battery $W$ is equal to the total charge delivered multiplied by the voltage:

$$W = Q \cdot V$$

The charge $Q$ stored in the capacitor is:

$$Q = C V$$

So, the work done becomes:

$$W = C V \cdot V = C V^2$$

3. Ratio of Energy Stored to Work Done

Now, the ratio of the energy stored in the capacitor to the work done by the battery is:

$$\text{Ratio} = \frac{U}{W} = \frac{\frac{1}{2} C V^2}{C V^2} = \frac{1}{2}$$

Final Answer:

The ratio is $\frac{1}{2}$.