Velocity ($$v$$) and acceleration ($$a$$) in two systems of units 1 and 2 are related as $$v_2 = \frac{n}{m^2} v_1$$ and $$a_2 = \frac{a_1}{mn}$$ respectively. Here $$m$$ and $$n$$ are constants. The relations for distance and time in two systems respectively are

We are given the relations between velocity and acceleration in two systems of units: $$v_2 = \frac{n}{m^2} v_1$$ and $$a_2 = \frac{a_1}{mn}$$.

Velocity has dimensions $$[LT^{-1}]$$ and acceleration has dimensions $$[LT^{-2}]$$, so $$\frac{L_2}{T_2} = \frac{n}{m^2} \cdot \frac{L_1}{T_1}$$ ... (i) and $$\frac{L_2}{T_2^2} = \frac{1}{mn} \cdot \frac{L_1}{T_1^2}$$ ... (ii).

Dividing equation (i) by (ii) yields $$\frac{L_2/T_2}{L_2/T_2^2} = \frac{\frac{n}{m^2} \cdot \frac{L_1}{T_1}}{\frac{1}{mn} \cdot \frac{L_1}{T_1^2}}$$ and thus $$T_2 = \frac{n}{m^2} \times mn \times T_1 = \frac{n^2}{m} T_1$$.

Substituting back into equation (i) gives $$L_2 = \frac{n}{m^2} \cdot \frac{L_1 \cdot T_2}{T_1} = \frac{n}{m^2} \cdot L_1 \cdot \frac{n^2/m}{1} = \frac{n^3}{m^3} L_1$$.

Therefore, $$L_2 = \frac{n^3}{m^3} L_1$$ and $$T_2 = \frac{n^2}{m} T_1$$.

The correct answer is Option A.