Answer
See the detailed answer below.
Work Step by Step
We know that the discharge of a capacitor through a resistor is given by
$$Q=Q_0e^{-t/\tau}$$
Hence, the current is given
$$I=I_0e^{-t/\tau}$$
where $\tau=RC$,
$$I=I_0e^{-t/RC} $$
Thus,
$$\ln\left(\dfrac{I}{I_0}\right)=\dfrac{-t}{RC}$$
$$\ln\left(I\right)-\ln\left(I_0\right)=-\dfrac{t}{RC}$$
$$\ln\left(I\right) =\ln\left(I_0\right)-\dfrac{t}{RC}$$
We can consider this as a straight line formula $y=mx+b$ where
$\ln\left(I\right)=y$,
slope $=m=-\dfrac{1}{RC}$,
$x=t$, and
$b=\ln\left(I_0\right)$
Using the given data in the given table to draw the best-fit line of this formula and then find its slope to find the resistance of the resistor$R$.
$$R=\dfrac{-1}{C\cdot{\rm Slope}}\tag 1$$
Also the initial current of the capacitor is given by the $y$-intercept.
$$b=\ln\left(I_0\right)$$
Hence,
$$e^b=I_0\tag 2$$
From the graph below we can see that ${\rm Slope}=$, and $b=$.
Plug into (1) and (2) respectively.
$$\color{blue}{\bf [a]}$$
$$R=\dfrac{-1}{C\cdot{\rm Slope}} =\dfrac{-1}{(20\times 10^{-6})(-0.77)}$$
$$R=\color{red}{\bf 64.9}\;\rm k\Omega$$
$$\color{blue}{\bf [b]}$$
$$I_0=e^b=e^{-6.61}=\color{red}{\bf 1.35 }\;\rm mA $$
