As illustrated in the figure, each plate possesses a surface area...

Let the four plates be numbered 1, 2, 3, and 4 from left to right.

From the given circuit diagram:
Plate 1 is connected to the positive terminal of the battery, so its potential is $V_1 = 10 \text{ V}$.
Plate 3 is connected to the negative terminal of the battery, so its potential is $V_3 = 0 \text{ V}$.
Plates 2 and 4 are connected to each other by a wire, forming an isolated system. Let their common potential be $V_2 = V_4 = V'$.

The arrangement forms three identical capacitors between the consecutive pairs of plates: $C_{12}$, $C_{23}$, and $C_{34}$.
The capacitance of each capacitor is given by:
$C = \dfrac{\varepsilon_0 A}{d}$
Given $\varepsilon_0 A = 96 \times 10^{-12} \text{ F}\cdot\text{m}$ and $d = 4.0 \text{ mm} = 4 \times 10^{-3} \text{ m}$.
$C = \dfrac{96 \times 10^{-12}}{4 \times 10^{-3}} = 24 \times 10^{-9} \text{ F} = 24 \text{ nF}$

Since Plates 2 and 4 form an isolated system, the net charge on them is zero:
$q_2 + q_4 = 0$
The charge on Plate 2 is the sum of charges on its left and right surfaces:
$q_2 = C(V_2 - V_1) + C(V_2 - V_3) = C(V' - 10) + C(V' - 0)$
The charge on Plate 4 is on its left surface (facing Plate 3):
$q_4 = C(V_4 - V_3) = C(V' - 0)$

Substituting these into the charge conservation equation:
$C(V' - 10) + CV' + CV' = 0$
$3V' - 10 = 0 \implies V' = \dfrac{10}{3} \text{ V}$

Now, we calculate the charge on the plates connected to the battery (Plate 1 and Plate 3).
Charge on Plate 1:
$q_1 = C(V_1 - V_2) = C \left(10 - \dfrac{10}{3}\right) =C \left(\dfrac{20}{3}\right)<br>$q_1 = 24 $\text{ nF} \times \dfrac{20}{3} \text{ V} =$160 $\text{ nC} =$0.16 \, \$mu\text{C}<br><br>Charge on Plate 3:<br>$q_3 = C(V_3 - V_2) + C(V_3 - V_4) = C \left(0 - \dfrac{10}{3}\right) +C \left(0 -\dfrac{10}{3}\right) = --$C\left(\dfrac{20}{3}\right)<br>$q_3 = -24 $\text{ nF} \times \dfrac{20}{3} \text{ V} = -$160 $\text{ nC} = -$0.16 \, \mu\text{C}$<br><br>The magnitude of the charge delivered by the battery to each of the plates linked to it is$0.16 \, \$mu\text{C}.<br><br>Answer:$0.16 \, \mu\text{C}