Let the upward direction be positive. The acceleration of the dropped coin relative to the ground is $a_{c,g} = -g$. The acceleration of the coin relative to the elevator is given as $a_{c,e} = -a_{rel}$.

The relationship for relative acceleration is $a_{c,g} = a_{c,e} + a_{e,g}$, where $a_{e,g}$ is the elevator's acceleration relative to the ground. We can solve for the elevator's acceleration:

$$a_{e,g} = a_{c,g} - a_{c,e} = (-g) - (-a_{rel}) = a_{rel} - g$$

Now, apply Newton's second law to the elevator system of mass $M$. The forces are the upward tension $T$ and the downward weight $Mg$.

$$\sum F = T - Mg = M a_{e,g}$$

Substitute the expression for the elevator's acceleration:

$$T = Mg + M(a_{rel} - g)$$ $$T = Mg + M a_{rel} - Mg$$ $$T = M a_{rel}$$

The tension is the product of the total mass and the apparent acceleration due to gravity inside the cab.