kengloong,
One way is to charge the capacitor to a known voltage. Then connect a known resistance (R) across the capacitor. The load resistance must be much less than the input resistance of the multimeter. Measure the time (t) required for the capacitor to discharge to 36.8% of its original value. The capacitance is equal to
t/R. This method is not practical for very small values of capacitance. Also, you must choose the value of R so that the response time of the multimeter is small compared to t.
Regards,
Kral