To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \beta t$, where $I_0 = 20$ A (initial current) and $\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:

$Q = \int_{0}^{T} I(t) \, dt$

Where $I(t) = I_0 + \beta t$ and $T = 20$ s. Substituting the given values:

$Q = \int_{0}^{20} (20 + 3t) \, dt$

This integral can be solved in two parts:

1. The integral of the constant term $20$:

$\int 20 \, dt = 20t$

2. The integral of the linear term $3t$:

$\int 3t \, dt = \frac{3}{2}t^2$

Thus, combining these and evaluating from $0$ to $20$ seconds:

$Q = \left[20t + \frac{3}{2}t^2\right]_{0}^{20}$

$Q = \left[20(20) + \frac{3}{2}(20)^2\right] - \left[20(0) + \frac{3}{2}(0)^2\right]$

$Q = 400 + 600 = 1000$ C

Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.