Answer
$C = 1.61\times 10^{-10}~F$
Work Step by Step
We can find the time constant $\tau$:
$\frac{q}{C} = \mathscr{E}~(1-e^{-t/\tau})$
$\frac{q}{C} = 12.0~V~(1-e^{-1.30~\mu s/\tau}) = 5.00~V$
$1-e^{-1.30~\mu s/\tau} = \frac{5.00}{12.0}$
$e^{-1.30~\mu s/\tau} = 1-\frac{5.00}{12.0}$
$e^{-1.30~\mu s/\tau} = \frac{7.00}{12.0}$
$e^{1.30~\mu s/\tau} = \frac{12.0}{7.00}$
$\frac{1.30~\mu s}{\tau} = ln(\frac{12.0}{7.00})$
$\tau = \frac{1.30~\mu s}{ln(\frac{12.0}{7.00})}$
$\tau = 2.41~\mu s$
We can find the capacitance $C$:
$RC = \tau$
$C = \frac{\tau}{R}$
$C = \frac{2.41~\mu s}{15.0~k \Omega}$
$C = \frac{2.41\times 10^{-6}~s}{1.50\times 10^4~\Omega}$
$C = 1.61\times 10^{-10}~F$