To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:

$$ I \propto \frac{1}{r^2} $$

This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $ (2r)^2 = 4r^2 $.

The initial intensity given at the origin (source) is:

$$ I_0 = 16 \times 10^{-8} \mathrm{~Wm}^{-2} $$

Let's call $ I_1 $ the intensity at $ r = 2 \text{ m} $ and $ I_2 $ the intensity at $ r = 4 \text{ m} $. Using the inverse square law, we can write:

$$ I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4} $$

and

$$ I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16} $$

Now substitute the given value for $ I_0 $ to find $ I_1 $ and $ I_2 $:

$$ I_1 = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~Wm}^{-2} $$

$$ I_2 = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~Wm}^{-2} $$

Now to find the difference in intensity (magnitude only) between the two points, we subtract $ I_2 $ from $ I_1 $:

$$ \Delta I = | I_1 - I_2 | $$

$$ \Delta I = | 4 \times 10^{-8} - 1 \times 10^{-8} | $$

$$ \Delta I = 3 \times 10^{-8} \mathrm{~Wm}^{-2} $$

So the difference in intensity (magnitude only) at the two points is:

$$ 3 \times 10^{-8} \mathrm{~Wm}^{-2} $$