If b is the mean proportion between a and c, then $\dfrac{a^2 − b^2 + c^2}{a^{-2} − b^{-2} + c^{-2}}$ is equal to:

1. a⁴

2. b⁴

3. a²

4. b²

Since b is the mean proportion between a and c,

$\therefore \dfrac{a}{b} = \dfrac{b}{c}$

$\Rightarrow b^2 = ac$

Solving, $\Rightarrow \dfrac{a^2 - b^2 + c^2}{a^{-2} - b^{-2} + c^{-2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{1}{a^{2}} - \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2c^2}{a^2b^2c^2} - \dfrac{a^2c^2}{a^2b^2c^2} + \dfrac{a^2b^2}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2c^2}{a^2b^2c^2} - \dfrac{(b^2)^2}{a^2b^2c^2} + \dfrac{a^2b^2}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2(c^2 - b^2 + a^2)}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{(c^2 - b^2 + a^2)}{a^2c^2}} \\[1em] \Rightarrow a^2 - b^2 + c^2 \times {\dfrac{a^2c^2}{(c^2 - b^2 + a^2)}} \\[1em] \Rightarrow a^2c^2 = (b^2)^2 = b^4.$

Hence, option 2 is the correct option.