Answer
${\bf 72.8}\;\rm \Omega$
Work Step by Step
We know that the discharge of a capacitor through a resistor is given by
$$Q=Q_0e^{-t/\tau}$$
where $\tau=RC$,
$$Q=Q_0e^{-t/RC}$$
Recalling that $\Delta V_C=Q/C$, so $Q_0=C\Delta V_0$, and $Q=C\Delta V$
Hence,
$$\color{red}{\bf\not} C\Delta V=\color{red}{\bf\not} C\Delta V_0e^{-t/RC}$$
$$ \Delta V= Delta V_0e^{-t/RC}$$
Hence,
$$\ln\left[ \dfrac{\Delta V}{\Delta V_0} \right]=\dfrac{-t}{RC}$$
Therefore,
$$R =\dfrac{-t}{ C\ln\left[ \dfrac{\Delta V}{\Delta V_0} \right]}$$
Plug the given; where we can see from the given graph that at $t=0$ ms, $\Delta V_0=30$ V, and at $t=4$ ms $\Delta V=10$ V.
$$R =\dfrac{-(4\times 10^{-3})}{ (50\times 10^{-6})\ln\left[ \dfrac{10}{30} \right]}$$
$$R=\color{red}{\bf 72.8}\;\rm \Omega$$
