Answer:
[tex]R=803k\Omega[/tex]
Explanation:
We have the following information,
[tex]V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s[/tex]
We apply the equation for capacitor charging the voltage across it,
[tex]V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}[/tex]
Replacing values,
[tex]R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}[/tex]
[tex]R=803k\Omega[/tex]