Answer
${\bf 5.2}\;\rm k\Omega$
Work Step by Step
We know that the discharge of a capacitor through a resistor is given by
$$Q=Q_0e^{-t/\tau}$$
When it is connected to the patient's chest it loses 0.9 of $Q_0$, so that $Q=0.9Q_0$
$$0.9Q_0=Q_0e^{-t/\tau}$$
where $\tau=RC$,
$$0.9 = e^{-t/RC}$$
Hence,
$$\ln(0.9)=\dfrac{-t}{RC}$$
Solving for $R$;
$$R=\dfrac{-t}{\ln(0.9) \cdot C}$$
Plug the given;
$$R=\dfrac{-(40\times 10^{-3})}{\ln(0.9) \cdot (150\times 10^{-6})}$$
$$R=\color{red}{\bf 5.2}\;\rm k\Omega$$
