The area is the sum of the areas of a rectangle and a triangle. These are found separately, first the rectnagle

$$
begin{align*}
A_{text{rectangle}} = p_{text{i}} (V_{text{f}} – V_{text{i}})
end{align*}
$$

Plugging in the numbers:

$$
begin{align*}
A_{text{rectangle}} = 110text{ kPa} cdot (0.62text{ m}^3 – 0.40text{ m}^3) = 2.42 cdot 10^{4}text{ J}
end{align*}
$$

and now the triangle:

$$
begin{align*}
A_{text{triangle}} = frac{1}{2} left(p_{text{f}} – p_{text{i}} right) left(V_{text{f}} – V_{text{i}} right)
end{align*}
$$

Plugging in the numbers:

$$
begin{align*}
A_{text{triangle}} &= frac{1}{2} left(260text{ kPa} – 110text{kPa} right) left(0.62text{ m}^3 – 0.40text{ m}^3 right) \
A_{text{triangle}} &= frac{1}{2} cdot 150text{ kPa} cdot 0.22text{ m}^3 = 1.65 cdot 10^{4}text{ J}
end{align*}
$$

The total work is:

$$
begin{align*}
W &= A_{text{rectangle}} + A_{text{triangle}} \
W &= 2.42 cdot 10^{4}text{ J} + 1.65 cdot 10^{4}text{ J} \
W &= 4.07 cdot 10^{4}text{ J}
end{align*}
$$

For process A we see that the area is a rectangle with sides $4text{ kPa}$ and $1text{ m}^3$. Hence the work done in process A is:

$$
begin{align*}
W_{text{A}} = 4text{ kPa} cdot 1text{ m}^3 = 4text{ J}
end{align*}
$$

For process B we see that the area is the sum of the areas of a rectangle with sides $1text{ kPa}$ and $3text{ m}^3$, a triangle with sides $2text{ kPa}$ and $2text{ m}^3$. Hence the work done in process B is:

$$
begin{align*}
W_{text{B}} &= 1text{ kPa} cdot 3text{ m}^3 + frac{1}{2} cdot 2text{ kPA} cdot 2text{ m}^3 \
W_{text{B}} &= 3text{ J} + 2text{ J} = 5text{ J}
end{align*}
$$

For process C we see that the area is a rectangle with sides $1text{ kPa}$ and $3text{ m}^3$. Hence the work done in process C is:

$$
begin{align*}
W_{text{C}} = 1text{ kPa} cdot 3text{ m}^3 = 3text{ J}
end{align*}
$$

Now to rank them in order of increasing work:

$$
begin{align*}
W_{text{C}} < W_{text{A}} < W_{text{B}}
end{align*}
$$

$$
begin{align*}
C < A < B
end{align*}
$$