**Calculation of Work Done**

To calculate the work done by the force, we will use the formula:

W = F * d * cos(theta)

where W is the work done, F is the force applied, d is the displacement, and theta is the angle between the force and the displacement vectors.

**Magnitude and Direction of Force**

The force is given as 20 N in the direction of 6i + 8j. To find the magnitude of the force, we can use the Pythagorean theorem:

|F| = sqrt((6^2)+(8^2))
|F| = sqrt(36+64)
|F| = sqrt(100)
|F| = 10 N

So, the magnitude of the force is 10 N.

To find the direction of the force, we can calculate the unit vector in the direction of 6i + 8j:

u = (6i + 8j)/sqrt((6^2)+(8^2))
u = (6i + 8j)/10
u = 0.6i + 0.8j

So, the direction of the force is 0.6i + 0.8j.

**Displacement Vector**

The displacement vector is given as the vector from point A (2,1,3) to point B (3,3,4):

d = (3-2)i + (3-1)j + (4-3)k
d = i + 2j + k

**Angle between Force and Displacement Vectors**

To find the angle between the force and displacement vectors, we can use the dot product:

cos(theta) = (F . d)/(|F| * |d|)
cos(theta) = ((10)(0.6i + 0.8j) . (i + 2j + k))/(10 * sqrt(6))
cos(theta) = (6 + 16 + 8)/(10 * sqrt(6))
cos(theta) = 3/5

So, the angle between the force and displacement vectors is:

theta = cos^-1(3/5)
theta = 53.13 degrees

**Calculation of Work Done**

Now, we can calculate the work done by the force:

W = (10)(i + 2j + k)(0.6i + 0.8j)cos(53.13)
W = (10)(0.6 + 1.6 + 0.8)(0.6)
W = 22.08 J

So, the work done by the force is approximately 22 J.

**Answer**

Therefore, the correct option is 1) 22.

C) 44 J