For the given input voltage waveform $V_{in}\left(t\right)$, the output waveform  $V_0\left(t\right)$, across the capacitor is correctly depicted by :

Time constant  $\tau =RC={10}^3\times 10\times {10}^{-9}={10}^{-5}=10\mu s$
 $V_0=\frac{q}{C}=v_{max}\left[1-e^{-t/\tau }\right]=5\left(1-e^{-t/10}\right)wheretis\mu s$
 $Att=5\mu s,V_0=5\left(1-e^{-1/2}\right)=2V$

$\begin{array}{l}For 5\mu s< t <10\mu s, the discharging of capacitor takes place\\ \text{}{V}_0=2\left(e^{-\frac{t}{\tau }}\right)\\ At t=10\mu s , Since discharging happened for duration of 5\mu s\text{}\\ V_0=2\left(e^{-\frac{5}{10}}\right)=1.21 V\\ For 10\mu s<t<15\mu s, again capacitor is in charging mode for duration of 5\mu s\text{}\\ So, at t=15\mu s, \\ V_0=1.21 +5\left(1-e^{-\frac{5}{10}}\right) =1.21+1.96=3.17 V\\ \end{array}$