Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know for a charging capacitor that
$$Q=Q_{\rm max}\left(1-e^{-t/RC}\right)$$
Recalling that $Q=C\Delta V_C$, so
$$C\Delta V_C=C(\Delta V_C)_{\rm max}\left(1-e^{-t/RC}\right)$$
where $(\Delta V_C)_{\rm max}=\varepsilon$, see the given graph.
$$ \Delta V_C=\varepsilon \left(1-e^{-t/RC}\right)$$
Let's assume that at $t=0$, $ \Delta V_C=V_{\rm off}$ and at the end of the period of oscillation, $ \Delta V_C=V_{\rm on}$; where the period is given by
$$T=t_{\rm on}-t_{\rm off}\tag 1$$
Hence
$$ V_{\rm off}=\varepsilon \left(1-e^{-t_{\rm off}/RC}\right)$$
Solving for $t_{\rm off}$;
$$ V_{\rm off}=\varepsilon -\varepsilon e^{-t_{\rm off}/RC} $$
$$\dfrac{\varepsilon -V_{\rm off}}{\varepsilon}= e^{-t_{\rm off}/RC} $$
Thus,
$$\ln\left[\dfrac{\varepsilon -V_{\rm off}}{\varepsilon}\right]= \dfrac{-t_{\rm off}}{RC} $$
$$t_{\rm off} =-RC\ln\left[\dfrac{\varepsilon -V_{\rm off}}{\varepsilon}\right] $$
$$t_{\rm off} = RC\ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm off}}\right] \tag 2$$
By the same approach,
$$t_{\rm on} = RC\ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm on}}\right] \tag 3$$
Plug (2) and (3) into (1),
$$T=RC\ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm on}}\right]-RC\ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm off}}\right]$$
$$T=RC\left(\ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm on}}\right]- \ln\left[\dfrac{\varepsilon}{\varepsilon -V_{\rm off}}\right]\right)$$
$$T=RC \ln\left[\dfrac{\dfrac{\varepsilon}{\varepsilon -V_{\rm on}}}{\dfrac{\varepsilon}{\varepsilon -V_{\rm off}}}\right] $$
$$\boxed{T=RC \ln\left[\dfrac{ \varepsilon -V_{\rm off}}{\varepsilon -V_{\rm on}}\right] }$$
$$\color{blue}{\bf [a]}$$
Recalling that $T=1/f$, and using the boxed formula above,
$$\dfrac{1}{f}=RC \ln\left[\dfrac{ \varepsilon -V_{\rm off}}{\varepsilon -V_{\rm on}}\right]$$
Hence,
$$R=\dfrac{1}{fC \ln\left[\dfrac{ \varepsilon -V_{\rm off}}{\varepsilon -V_{\rm on}}\right]}$$
Plug the given;
$$R=\dfrac{1}{(10)(10\times 10^{-6})\ln\left[\dfrac{90 -20}{90 -80}\right]}$$
$$R=\color{red}{\bf 5.12}\;\rm k\Omega$$
