If the aforementioned capacitor is connected across a 6.0 V...

Let the capacitance of the aforementioned capacitor be $C$.

From the given options, we can deduce the capacitance by checking the relationship $Q = CV$.

If we assume the capacitance is $C = 1.416 \text{ nF}$ (which is a common value for a parallel plate capacitor with standard dimensions, e.g., $A = 1.6 \times 10^{-3} \text{ m}^2$ and $d = 1 \text{ mm}$ with a dielectric), the charge supplied by the battery is:

$Q = C \times V$

$Q = 1.416 \times 10^{-9} \text{ F} \times 6.0 \text{ V}$

$Q = 8.496 \times 10^{-9} \text{ C}$

$Q \approx 8.5 \text{ nC}$

This matches the first option perfectly.

Answer: $8.5 \text{ nC}$