B

Step 1: Understand the setup.
We have a spherical capacitor with inner radius $a$ and outer radius $b$. The dielectric permittivity varies as $\varepsilon(r) = \frac{\alpha}{r}$, meaning it changes as we move from the inner to the outer sphere.

Step 2: Use the formula for the capacitance of a capacitor.
The capacitance $C$ can be derived from the relationship between charge $Q$, voltage $V$, and capacitance, which gives $C = \frac{Q}{V}$. In the case of a spherical capacitor, we can express the voltage as an integral involving the electric field $E$.

Step 3: Consider the electric field. The electric field $E$ can be found from $E = -\frac{dV}{dr}$. Using Gauss's law for a spherical shell, we have:
$$ Q = \varepsilon(r) \cdot 4\pi r^2 E $$
Since $E = \frac{Q}{4\pi \varepsilon(r) r^2}$, we can rewrite the voltage as:
$$ V = -\int_a^b E dr = -\int_a^b \frac{Q}{4\pi \varepsilon(r) r^2} dr $$

Step 4: Substitute $\varepsilon(r) = \frac{\alpha}{r}$ into the equation:
$$ V = -\int_a^b \frac{Q}{4\pi \left(\frac{\alpha}{r}\right) r^2} dr = -\int_a^b \frac{Q r}{4\pi \alpha} dr $$
Integrating yields:
$$ V = -\frac{Q}{4\pi \alpha} \left[ \frac{r^2}{2} \right]_a^b = -\frac{Q}{4\pi \alpha} \left( \frac{b^2 - a^2}{2} \right) $$

Step 5: Substitute $V$ back into the capacitance formula:
$$ C = \frac{Q}{V} = \frac{Q}{-\frac{Q}{4\pi \alpha} \left( \frac{b^2 - a^2}{2} \right)} = -\frac{4\pi \alpha}{ \frac{b^2 - a^2}{2}} $$

Step 6: Simplifying gives:
$$ C = \frac{8\pi \alpha}{b^2 - a^2} $$
Thus, the capacitance of the spherical capacitor filled with the given dielectric is option B. Therefore, the correct answer is:
Therefore, B.