A child stands on a weighing machine inside a lift. When the lift is going down with acceleration $\frac{g}{3}$, the machine shows a reading 20 N. When the lift goes upwards with acceleration $\frac{g}{3}$, the reading would be ( $\mathrm{g}=$ gravitational acceleration)

Mass and apparent weight
Let the mass of the child be $m$, and $R$ be the reading on the scale, which equals the normal force exerted on the child.

The apparent weight changes when the elevator accelerates. Applying Newton's second law, $\Sigma F = ma$:

Elevator descending with acceleration $a = \frac{g}{3}$
The forces are the gravitational force $mg$ downward and the normal force $R_1$ upward. Since the acceleration is downward, $mg - R_1 = ma$. Given $R_1 = 20\text{ N}$ and $a = \frac{g}{3}$, we have:

$mg - 20 = \frac{mg}{3}$
$\frac{2mg}{3} = 20$
$mg = 30\text{ N}$

The actual weight is $30\text{ N}$.

Elevator ascending with acceleration $a = \frac{g}{3}$
Now, the net force is upward: $R_2 - mg = ma$. With $mg = 30\text{ N}$ and $a = \frac{g}{3}$, substitute:

$R_2 = 30 + \frac{mg}{3} = 30 + 10 = 40\text{ N}$

The scale reads $\boxed{40\text{ N}}$ when the elevator ascends with acceleration $\frac{g}{3}$.