Figure shows a network of five capacitors connected to a supply voltage 'V'. The equivalent capacitance and the energy stored in the network is respectively

Let the positive terminal of the battery be connected to node A and the negative terminal to node C. Let the top wire be node B.

From the given circuit diagram, we can identify the connections of each capacitor:
The top $3C$ capacitor is connected between nodes A and B.
The vertical $3C$ capacitor is connected between nodes A and B.
The middle horizontal $2C$ capacitor is connected between nodes A and C.
The vertical $1C$ capacitor is connected between nodes B and C.
The rightmost vertical $2C$ capacitor is connected between nodes B and C.

The two $3C$ capacitors are in parallel. Their equivalent capacitance is:
$C_{AB} = 3C + 3C = 6C$

The $1C$ and $2C$ capacitors are in parallel. Their equivalent capacitance is:
$C_{BC} = 1C + 2C = 3C$

The equivalent capacitors $C_{AB}$ and $C_{BC}$ are in series. Their equivalent capacitance is:
$C_{series} = \dfrac{C_{AB} \times C_{BC}}{C_{AB} + C_{BC}} = \dfrac{6C \times 3C}{6C + 3C} = 2C$

This series combination is in parallel with the middle $2C$ capacitor. The total equivalent capacitance of the network is:
$C_{eq} = C_{series} + 2C = 2C + 2C = 4C$

The total energy stored in the network is given by:
$U = \dfrac{1}{2} C_{eq} V^2 = \dfrac{1}{2} (4C) V^2 = 2CV^2$

Answer: $4C$ , $2CV^2$