Multimeter input impedance and its effect on the measurement of charged capacitor's voltage?
For a given time constant \$\tau\$, the required value of input resistance is

$$R = \frac{1}{\tau C}$$

For example, for a 1 second time constant and 15pF capacitance, the meter input resistance would be

$$R = \frac{1}{15 \cdot 10^{-12}} = 66.7 \,\mathrm G\Omega$$

Now, that's an enormous input resistance and the capacitor will still discharge in about 5 seconds.

Without further information on what you would like to accomplish, it's difficult to advise you on how to proceed.  The calculation above is simply to give you an idea of what you're up against.