An assembly comprises three conducting concentric spherical shells...

The given assembly can be considered as two spherical capacitors connected in series because the middle shell is isolated and uncharged.

The first capacitor $C_1$ is formed by the inner shell of radius $a$ and the middle shell of radius $b$. Its capacitance is:
$C_1 = \dfrac{4\pi\varepsilon_0 a b}{b - a}$

The second capacitor $C_2$ is formed by the middle shell of radius $b$ and the outer shell of radius $c$. Its capacitance is:
$C_2 = \dfrac{4\pi\varepsilon_0 b c}{c - b}$

The equivalent capacitance $C_{eq}$ for the series combination is given by:
$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$

Substituting the values of $C_1$ and $C_2$:
$\dfrac{1}{C_{eq}} = \dfrac{b - a}{4\pi\varepsilon_0 a b} + \dfrac{c - b}{4\pi\varepsilon_0 b c}$

\dfrac{1}{C_{eq}} = \dfrac{1}{4\pi\varepsilon_0} \left(\dfrac{1}{a} - \dfrac{1}{b} + \dfrac{1}{b} - \dfrac{1}{c} \right)

\dfrac{1}{C_{eq}} = \dfrac{1}{4\pi\varepsilon_0} \left( \dfrac{1}{a} - \dfrac{1}{c} \right) = \dfrac{c - a}{4\pi\varepsilon_0 a c}

Therefore, the equivalent capacitance is:
$C_{eq} =$\dfrac{4\pi\varepsilon_0 a c}{c - a}