Given charge as a function of time

$Q(t)=\alpha t-\beta t^2+\gamma t^3$

Current is the time derivative of charge

$I(t)=\frac{dQ}{dt}$

So,

$I(t)=\alpha-2\beta t+3\gamma t^2$

To find the minimum value of current

Since $I(t)$ is a quadratic in $t$:

$I(t)=3\gamma t^2-2\beta t+\alpha$

Its minimum occurs at the vertex, provided the coefficient of $t^2$ is positive ($\gamma>0$).

For a quadratic $at^2+bt+c$, the vertex is at

$t=-\frac{b}{2a}$

Here,

$a=3\gamma, \quad b=-2\beta$

Hence,

$t_{\min}=\frac{2\beta}{2\cdot 3\gamma}=\frac{\beta}{3\gamma}$

Substitute this value into $I(t)$

$I_{\min}=\alpha-2\beta\left(\frac{\beta}{3\gamma}\right)+3\gamma\left(\frac{\beta}{3\gamma}\right)^2$

Simplify term by term:

$I_{\min}=\alpha-\frac{2\beta^2}{3\gamma}+3\gamma\cdot \frac{\beta^2}{9\gamma^2}$

$I_{\min}=\alpha-\frac{2\beta^2}{3\gamma}+\frac{\beta^2}{3\gamma}$

$I_{\min}=\alpha-\frac{\beta^2}{3\gamma}$

Match with the options

This corresponds to:

$\boxed{\alpha-\frac{\beta^2}{3\gamma}}$

So, the correct option is C.