As shown in the figure, a gas can be taken from state A to state B...

From the given $V-p$ graph, the coordinates of the states are:
$p_A = 10 \text{ kPa}$, $V_A = 10 \text{ cc}$
$p_B = 30 \text{ kPa}$, $V_B = 25 \text{ cc}$
$p_C = 30 \text{ kPa}$, $V_C = 10 \text{ cc}$
$p_D = 10 \text{ kPa}$, $V_D = 25 \text{ cc}$

The work done by the gas is given by $W = \int p \, dV$, which is the area under the curve projected onto the volume ($V$) axis.

For path $AB$:
The work done is the area of the trapezium formed by the line $AB$ and the $V$-axis.
$W_{AB} = \dfrac{1}{2} (p_A + p_B) (V_B - V_A)$
$W_{AB} = \dfrac{1}{2} (10 + 30) \times 10^3 \times (25 - 10) \times 10^{-6}$
$W_{AB} = \dfrac{1}{2} \times 40 \times 10^3 \times 15 \times 10^{-6} = 0.30 \text{ J}$

For path $ACB$:
The process $AC$ is isochoric ($V = \text{constant}$), so $W_{AC} = 0$.
The process $CB$ is isobaric ($p = \text{constant}$), so $W_{CB} = p_C (V_B - V_C)$.
$W_{CB} = 30 \times 10^3 \times (25 - 10) \times 10^{-6} = 30 \times 15 \times 10^{-3} = 0.45 \text{ J}$
$W_{ACB} = W_{AC} + W_{CB} = 0 + 0.45 = 0.45 \text{ J}$

For path $ADB$:
The process $AD$ is isobaric ($p = \text{constant}$), so $W_{AD} = p_A (V_D - V_A)$.
$W_{AD} = 10 \times 10^3 \times (25 - 10) \times 10^{-6} = 10 \times 15 \times 10^{-3} = 0.15 \text{ J}$
The process $DB$ is isochoric ($V = \text{constant}$), so $W_{DB} = 0$.
$W_{ADB} = W_{AD} + W_{DB} = 0.15 + 0 = 0.15 \text{ J}$

The work done in paths $AB$, $ACB$, and $ADB$ are $0.30 \text{ J}$, $0.45 \text{ J}$, and $0.15 \text{ J}$ respectively.

Answer: $0.30\text{ J}$, $0.45\text{ J}$, and $0.15\text{ J}$