Determine the reading of the spring balance depicted in the figure....

The elevator is accelerating upwards with an acceleration $a = \dfrac{g}{10}$.

The effective acceleration due to gravity inside the elevator is:
$g_{\text{eff}} = g + a = g + \dfrac{g}{10} = \dfrac{11g}{10}$

For the masses $m_1 = 1.5 \text{ kg}$ and $m_2 = 3.0 \text{ kg}$ connected over the pulley, the tension $T$ in the string is given by:
$T = \dfrac{2m_1m_2}{m_1 + m_2} g_{\text{eff}}$

Substituting the given values:
$T = \dfrac{2 \times 1.5 \times 3.0}{1.5 + 3.0} \times \dfrac{11g}{10}$

$T = \dfrac{9.0}{4.5} \times \dfrac{11g}{10} = 2 \times \dfrac{11g}{10} = 2.2g$

The downward force exerted on the pulley, which is measured by the spring balance, is $2T$:
$F = 2T = 2 \times 2.2g = 4.4g$

Since the spring balance is calibrated in kg, its reading is the measured force divided by the true acceleration due to gravity $g$:
$\text{Reading} = \dfrac{F}{g} = \dfrac{4.4g}{g} = 4.4 \text{ kg}$