= Solution
Put $\omega_0^2=1/(LC)$. The initial conditions give
$$
q(t)=Qe^{-\gamma t}\left(\cos\omega_dt+\frac\gamma{\omega_d}\sin\omega_dt\right),\qquad V_C(t)=V_0e^{-\gamma t}\left(\cos\omega_dt+\frac\gamma{\omega_d}\sin\omega_dt\right).
$$
The printed <electric current> arrow points toward the positively labelled <capacitor> plate, so its signed <electric current> is $I=\dot q$, not the positive discharge magnitude. Hence
$$
\boxed{I(t)=-\frac{V_0}{L\omega_d}e^{-\gamma t}\sin\omega_dt.}
$$
For the stated <electrical resistance>,
$$
\gamma=25\ \mathrm{s^{-1}},\quad \omega_0\simeq447.21\ \mathrm{s^{-1}},\quad \omega_d\simeq446.51\ \mathrm{s^{-1}},\quad T_d\simeq14.07\ \mathrm{ms}.
$$
\b[The <voltage> starts at $6\ \mathrm V$ with zero slope; the signed <electric current> starts at zero and initially becomes negative.] Its initial slope is $\dot I(0)=-V_0/L=-6\ \mathrm{A\,s^{-1}}$. Both oscillate with an $e^{-25t}$ envelope and a $40\ \mathrm{ms}$ amplitude <time constant>. The phase difference is close to a quarter period, with the small damping correction displayed explicitly in $V_C$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-rlc-transient.png]
{title=Signed capacitor voltage and printed-arrow current, with the exchange and dissipation of circuit energy over three periods}
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