And also calculate the total change in the 150 microfarad capacitor

When the switch Z is closed at point X, the $100 \, \mu F$ capacitor (let's call it $C_1$) is connected in series with the $1.5 \, k\Omega$ resistor and the $3.00 \, V$ battery. The $150 \, \mu F$ capacitor ($C_2$) is not part of this circuit.
The problem states that $C_1$ is fully charged. When a capacitor is fully charged in a DC circuit, the voltage across it equals the source voltage, and no current flows through the resistor.
The initial charge on $C_1$ is $Q_{C1, initial} = C_1 V_{battery}$.

$$Q_{C1, initial} = (100 \times 10^{-6} \, F) \times (3.00 \, V) = 300 \times 10^{-6} \, C = 300 \, \mu C$$

The $150 \, \mu F$ capacitor ($C_2$) is initially uncharged, so its initial charge is $Q_{C2, initial} = 0 \, \mu C$.

When Z is moved to Y, the $100 \, \mu F$ capacitor ($C_1$) is now connected in parallel with the $150 \, \mu F$ capacitor ($C_2$). The battery and resistor are no longer part of this closed loop.
Since the capacitors are connected in parallel, they will share the total charge until they reach a common voltage. The total charge in this isolated system is conserved.
Total initial charge $Q_{total} = Q_{C1, initial} + Q_{C2, initial} = 300 \, \mu C + 0 \, \mu C = 300 \, \mu C$.

The equivalent capacitance of $C_1$ and $C_2$ in parallel is $C_{eq} = C_1 + C_2$.

$$C_{eq} = 100 \, \mu F + 150 \, \mu F = 250 \, \mu F$$

The final common voltage across both capacitors, $V_{final}$, can be found using $Q_{total} = C_{eq} V_{final}$.

$$V_{final} = \frac{Q_{total}}{C_{eq}} = \frac{300 \, \mu C}{250 \, \mu F} = \frac{300 \times 10^{-6} \, C}{250 \times 10^{-6} \, F} = 1.2 \, V$$

The question asks for the total change in the $150 \, \mu F$ capacitor, which refers to the change in its charge.
The final charge on $C_2$ ($150 \, \mu F$) is $Q_{C2, final} = C_2 V_{final}$.

$$Q_{C2, final} = (150 \times 10^{-6} \, F) \times (1.2 \, V) = 180 \times 10^{-6} \, C = 180 \, \mu C$$

The change in charge on $C_2$ is $\Delta Q_{C2} = Q_{C2, final} - Q_{C2, initial}$.

$$\Delta Q_{C2} = 180 \, \mu C - 0 \, \mu C = 180 \, \mu C$$

The total change in the $150 \, \mu F$ capacitor is $\boxed{180 \, \mu C}$.