Each side of a box made of metal sheet in cubic shape is $$a$$ at room temperature $$T$$, the coefficient of linear expansion of the metal sheet is $$\alpha$$. The metal sheet is heated uniformly, by a small temperature $$\Delta T$$, so that its new temperature is $$T + \Delta T$$. Calculate the increase in the volume of the metal box.

We have a cubic metal box with each side of length $$a$$ at room temperature $$T$$. The coefficient of linear expansion of the metal sheet is $$\alpha$$. The temperature is increased by $$\Delta T$$.

The initial volume of the cube is $$V_0 = a^3$$.

When the temperature increases by $$\Delta T$$, each side of the cube expands. The new length of each side is $$a' = a(1 + \alpha \Delta T)$$.

The new volume is $$V = (a')^3 = a^3(1 + \alpha \Delta T)^3$$.

Since $$\alpha \Delta T$$ is very small (the problem states "small temperature $$\Delta T$$"), we can use the binomial approximation $$(1 + x)^3 \approx 1 + 3x$$ for small $$x$$.

So $$V \approx a^3(1 + 3\alpha \Delta T) = a^3 + 3a^3\alpha \Delta T$$.

The increase in volume is $$\Delta V = V - V_0 = 3a^3\alpha \Delta T$$.

This result is consistent with the general formula for volume expansion: $$\Delta V = V_0 \gamma \Delta T$$, where the coefficient of volume expansion $$\gamma = 3\alpha$$ for isotropic materials.

Hence, the correct answer is Option B.