B

Step 1: Analyze the forces acting on the pendulum bob at the mean position.
At the lowest point (mean position), the forces acting on the pendulum bob are:
1. Gravitational force (weight) acting downward: $W = mg$
2. Tension in the string ($T$) acting upward.

Given that the tension is 3 times the weight, we have:
$$T = 3W = 3mg$$

Step 2: Apply the centripetal force requirement at the mean position.
At the lowest point, the net force must equal the centripetal force required for circular motion:
$$T - W = \frac{mv^2}{L}$$
where $v$ is the speed of the pendulum bob and $L$ is the length of the pendulum.
So we write:
$$3mg - mg = \frac{mv^2}{L}$$
Simplifying gives:
$$2mg = \frac{mv^2}{L}$$
Cancelling $m$ (as long as $m \neq 0$) results in:
$$2gL = v^2$$
Hence, the speed at the mean position is: $v = \sqrt{2gL}$

Step 3: Use conservation of mechanical energy to find maximum displacement.
At maximum displacement (height $h$), all kinetic energy is converted to potential energy:
$$\frac{1}{2}mv^2 = mgh$$
Substituting for $v$ gives:
$$\frac{1}{2}m(2gL) = mgh$$
Cancelling $m$ from both sides results in:
$$gL = gh$$
This gives:
$$h = L$$

Since the maximum displacement relates to the vertical height ($h$) displaced, we look for the angle $ heta$:
$$\cos \theta = \frac{L - h}{L} = 0$$
This means $ heta = 90^\circ$.
The maximum displacement occurs when the pendulum swings to the point directly horizontal to the pivot point, across from the mean position.

The angle of maximum displacement is thus: $90^\circ$. Hence the correct angle, corresponding to displacement, is given by option B.