To solve this problem, we start by understanding the information given and how the polarization of light affects light intensity.

Initially, we have the ratio of the intensities at points A and B given as:

$$ r = \frac{I_A}{I_B} = 2 $$

This means:

$$ I_A = 2I_B $$

Now, we place two polaroids with their pass-axes at an angle of $45^\circ$ between them before point B. When unpolarized light passes through the first polaroid, it gets polarized, and its intensity is reduced to half of its original value:

$$ I_{B1} = \frac{1}{2} I_B $$

Here, $I_{B1}$ is the intensity of light after passing through the first polaroid.

Next, this polarized light passes through the second polaroid, which is at an angle of $45^\circ$ to the first one. According to Malus's law, the intensity of light after passing through the second polaroid is given by:

$$ I_{B2} = I_{B1} \cos^2(45^\circ) $$

We know that:

$$ \cos(45^\circ) = \frac{1}{\sqrt{2}} $$

Thus,

$$ I_{B2} = \frac{1}{2} I_B \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} I_B \cdot \frac{1}{2} = \frac{1}{4} I_B $$

Now, the new intensity at point B is $I_{B2}$. We need to calculate the new ratio $r'$ of the intensities at points A and B:

$$ r' = \frac{I_A}{I_{B2}} $$

Substituting the values $I_A = 2I_B$ and $I_{B2} = \frac{1}{4} I_B$, we get:

$$ r' = \frac{2 I_B}{\frac{1}{4} I_B} = 2 \times 4 = 8 $$

Therefore, the new value of $r$ will be:

$$ r' = 8 $$