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Voltage: A unit of measurement for electrical force (volts).

Capacitor: Device for storing electrical energy, consisting of two conductors in close proximity and insulated from each other.

The formula for the time constant,

$\mathrm{蟿}=\mathrm{RC}$

The formula for voltage of a charging capacitor as,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{蟿}}\right)$

The values of $\mathrm{C}=100pF\left(\frac{{10}^{-12}\mathrm{F}}{1pF}\right)=1.00脳{10}^{-10}\mathrm{F}$and$\mathrm{R}=75M惟\left(\frac{{10}^6惟}{1M惟}\right)=7.5脳{10}^7\mathrm{惟}$.

Firstly, use the formula for the time constant and substitute the given values,

$$\begin{gathered} \mathrm{蟿}=\mathrm{RC} \\ =\left(7.5脳{10}^7\mathrm{惟}\right)脳\left(1.00脳{10}^{-10}\mathrm{F}\right) \\ =7.5脳{10}^{-3}\mathrm{s}\left(\frac{1\mathrm{ms}}{{10}^{-3}\mathrm{s}}\right) \\ =7.5\mathrm{ms} \end{gathered}$$

Therefore, the time constant is$7.5\mathrm{ms}$

Apply the formula of the voltage of a charging capacitor to calculate time,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{蟿}}\right)$

The value of$V\left(t=T\right)=V_f$

And substitute ${\mathrm{V}}_{\mathrm{f}}=0.9{\mathrm{V}}_0$, therefore,

$$\begin{gathered} 0.9{\mathrm{V}}_0={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{T}/\mathrm{蟿}}\right) \\ 1-{\mathrm{e}}^{-\mathrm{T}/\mathrm{蟿}}=0.9 \\ -\frac{\mathrm{T}}{\mathrm{蟿}}=\ln \left(0.1\right) \\ \mathrm{T}=-\mathrm{蟿}\left[\ln \left(0.1\right)\right] \\ \mathrm{T}=0.0173\mathrm{s}\left(\frac{1\mathrm{ms}}{{10}^{-3}\mathrm{s}}\right) \\ \mathrm{T}=17.3\mathrm{s} \end{gathered}$$

Hence, the required time is $T=17.3\mathrm{ms}$.