B

Step 1: Calculate the charge on the capacitor after it is fully charged.
The capacitance $C$ is 200 µF = $200 \times 10^{-6}$ F and the emf $V$ is 10.0 V. The charge $Q$ on the capacitor when fully charged is given by the formula:
$$Q = C \times V$$
Substituting the known values:
$$Q = (200 \times 10^{-6}) \times 10 = 2000 \times 10^{-6} = 2 \times 10^{-3} ext{ C} = 2 ext{ mC}$$

Step 2: Determine the time constant of the circuit.
The time constant $\tau$ for an RC circuit is given by:
$$\tau = R \times C$$
where $R$ is the resistance (40 kΩ = $40 \times 10^{3}$ Ω). So,
$$\tau = (40 \times 10^{3}) \times (200 \times 10^{-6}) = 8$$ seconds.

Step 3: Calculate the voltage across the capacitor after charging for 16 seconds.
Since 16 seconds is twice the time constant, the capacitor will be very near its maximum charge. Using the equation for charging:
$$Q(t) = C \times V(1 - e^{-t/\tau})$$
Since $t = 16$ s and $\tau = 8$ s, we have:
$$Q(16) = C \times V(1 - e^{-16/8}) = Q_{max}(1 - e^{-2})$$
We know $e^{-2}$ is given as 0.135. So:
$$Q(16) = 2 \times 10^{-3} (1 - 0.135) = 2 \times 10^{-3} \times 0.865 = 1.73 \text{ mC}$$

Step 4: Determine the charge on the capacitor after disconnection.
After disconnecting the battery and connecting a thick wire, the voltage across the capacitor will immediately be 0. Thus, the charge will not be lost instantly. However, we need to evaluate the charge after another 16 seconds.

The charge decays according to the equation:
$$Q(t) = Q_0 e^{-t/\tau}$$
where $Q_0 = 1.73 ext{ mC}$ and $t = 16$ s. Thus:
$$Q(16) = 1.73 e^{-16/8} = 1.73 e^{-2}$$
Substituting $e^{-2} = 0.135$ gives:
$$Q(16) = 1.73 \times 0.135 = 0.23355 ext{ mC}$$
Finally, rounding off gives us approximately 0.234 mC.
Therefore, the charge on the capacitor after 16 seconds will be approximately 0.234 mC.