Calculate the energy stored in a wire with length 4 m, cross-sectional area 3e-6 m2, extension 1 mm, and Young's Modulus 2e11 Pa.

Given:
Length of wire, $l = 4 \text{ m}$
Cross-sectional area, $A = 3 \times 10^{-6} \text{ m}^2$
Extension, $e = 1 \text{ mm} = 1 \times 10^{-3} \text{ m}$
Young's Modulus, $E = 2 \times 10^{11} \text{ Pa}$

The energy stored (elastic potential energy) in a stretched wire is given by the formula:

$$U = \frac{1}{2} \frac{E A e^2}{l}$$

$$U = \frac{1}{2} \frac{(2 \times 10^{11}) \times (3 \times 10^{-6}) \times (1 \times 10^{-6})}{4}$$ $$U = \frac{2 \times 3 \times 10^{11 - 6 - 6}}{2 \times 4}$$ $$U = \frac{6 \times 10^{-1}}{8}$$ $$U = \frac{0.6}{8}$$ $$U = 0.075 \text{ J}$$

The energy stored in the wire is $0.075 \text{ J}$.

Comparing this with the given options:
A. 0.053J
B. 0.15J
C. 0.075J
D. 0.0075J

The correct option is C.

The final answer is $\boxed{\text{C. } \textbf{0.075J}}$