Calculate the acceleration for each of the three systems.

In System A, a mass $m$ is attached to one end of a string, and a force $F_1 = 2mg$ is applied to the other end. Assuming an ideal pulley and string, the tension $T$ throughout the string is equal to the applied force.

$$T = F_1 = 2mg$$

For the mass $m$, the forces acting are its weight $mg$ downwards and the tension $T$ upwards. Applying Newton's second law ($F_{net} = ma_1$):

$$T - mg = ma_1$$ $$2mg - mg = ma_1$$ $$mg = ma_1$$ $$a_1 = g$$

In System B, a mass $m$ is on one side and a mass $2m$ is on the other side. This is an Atwood machine. Let $T$ be the tension in the string and $a_2$ be the acceleration. We assume $2m$ moves down and $m$ moves up.
For mass $m$:

$$T - mg = ma_2 \quad (1)$$

For mass $2m$:

$$2mg - T = (2m)a_2 \quad (2)$$

Adding equations (1) and (2):

$$(T - mg) + (2mg - T) = ma_2 + 2ma_2$$ $$mg = 3ma_2$$ $$a_2 = \frac{mg}{3m}$$ $$a_2 = \frac{g}{3}$$

In System C, a mass $m$ is on the left, and a mass $m$ is on the right with an additional downward force $F_2 = mg$. Let $T$ be the tension and $a_3$ be the acceleration. We assume the right side moves down and the left side moves up.
For the left mass $m$:

$$T - mg = ma_3 \quad (1)$$

For the right mass $m$, the total downward force is its weight $mg$ plus the applied force $F_2 = mg$. So, the total downward force is $mg + mg = 2mg$.

$$2mg - T = ma_3 \quad (2)$$

Adding equations (1) and (2):

$$(T - mg) + (2mg - T) = ma_3 + ma_3$$ $$mg = 2ma_3$$ $$a_3 = \frac{mg}{2m}$$ $$a_3 = \frac{g}{2}$$

We have the accelerations:

$$a_1 = g$$ $$a_2 = \frac{g}{3}$$ $$a_3 = \frac{g}{2}$$

Comparing these values:
$g > \frac{g}{2} > \frac{g}{3}$
Therefore, $a_1 > a_3 > a_2$.

This matches option (b).

The final answer is $\text{(b)}$.