Answer
$9.2\times10^4 \Omega$.
Work Step by Step
The capacitor charges up to $75\%$ of its maximum value before it is discharged, and the cycle repeats. The desired time to reach that voltage is the interval between successive heartbeats.
$$t=\frac{60s}{72\;beats}=0.8333s$$
Use the equation for the voltage across a charging capacitor.
$$V_C=V_o(1-e^{-t/RC})$$
$$0.75 V_o =V_o(1-e^{-t/RC})$$
$$t=-RC \; ln(1-\frac{0.75}{1})$$
$$R=-\frac{t}{Cln(0.25)}=-\frac{0.8333s}{(6.5\times10^{-6}F)(-1.3863)}=9.2\times10^4\Omega$$
