B

To find the capacitance of the capacitor, we start with the formula for the charging of a capacitor in an RC (resistor-capacitor) circuit given by:
$$ V(t) = V_0 (1 - e^{-t/(RC)}) $$
where:
- $V(t)$ is the voltage across the capacitor at time $t$.
- $V_0$ is the supply voltage (12 V in this case).
- $R$ is the resistance (10 Ω).
- $C$ is the capacitance (which we want to find).

Since we need $V(t)$ at $t = 1 \mu s$, we can substitute the known values:
$$ 4.0 = 12 (1 - e^{-1 \mu s/(10 C)}) $$
Rearranging the equation yields:
$$ e^{-1 \mu s/(10 C)} = 1 - \frac{4.0}{12} $$
$$ e^{-1 \mu s/(10 C)} = \frac{8}{12} = \frac{2}{3} $$
Taking the natural logarithm of both sides, we have:
$$ -\frac{1 \mu s}{10 C} = \ln \left( \frac{2}{3} \right) $$
Thus, we can express $C$ in terms of known constants:
$$ C = -\frac{1 \mu s}{10 \ln \left( \frac{2}{3} \right)} $$
Plugging in the value of $\ln \left( \frac{2}{3} \right)$ (given) as $-0.4055$ gives us:
$$ C = -\frac{1 \times 10^{-6}}{10 (-0.4055)} \approx 2.46 \times 10^{-7} F = 246 nF $$
The closest answer is option B: 240 nF.