Step-by-step Solution:

1. Identify the given formula and variables: The formula for Young's modulus (Y) is given as: $Y = {{4MLg} \over {\pi l{d^2}}}$ The given measured values are:

   • Length of the wire, $L = 2$ m
   • Diameter of the wire, $d = 0.5$ mm
   • Mass, $M = 2.5$ kg
   • Extension in length, $l = 0.25$ mm

2. Determine the formula for maximum probable error: The maximum probable relative error in the measurement of Y can be found by taking the logarithm and differentiating the formula, or by using the standard formula for propagation of errors. Assuming M, L, and g are measured with high precision or are constants for this problem, the error in Y primarily depends on the errors in measuring $l$ and $d$. The relative error in Y is given by: ${\Delta Y \over Y} = {\Delta l \over l} + 2{{\Delta d} \over d}$ The terms on the right side represent the contributions to the total relative error from the measurements of $l$ and $d$ respectively.

   • Contribution due to error in $l$: $E_l = {\Delta l \over l}$
   • Contribution due to error in $d$: $E_d = 2{{\Delta d} \over d}$

3. Calculate the least count of the measuring instruments: Both diameter ($d$) and extension ($l$) are measured using instruments (a screw gauge and a micrometer) with the same specifications:

   • Pitch = 0.5 mm
   • Number of divisions on the circular scale = 100 The least count (LC) of these instruments is: $LC = {{\text{Pitch}} \over {\text{Number of divisions}}} = {{0.5 \text{ mm}} \over {100}} = 0.005 \text{ mm}$ The least count represents the maximum possible error (or uncertainty) in a single measurement. Therefore, we take:
   • Error in measuring $l$, $\Delta l = LC = 0.005$ mm
   • Error in measuring $d$, $\Delta d = LC = 0.005$ mm

4. Calculate the contributions to the error from $l$ and $d$: Now, we can calculate the numerical values of the contributions $E_l$ and $E_d$.

   • Contribution from extension ($l$): $E_l = {\Delta l \over l} = {{0.005 \text{ mm}} \over {0.25 \text{ mm}}} = {{5 \times 10^{-3}} \over {25 \times 10^{-2}}} = {5 \over 250} = {1 \over 50} = 0.02$

   • Contribution from diameter ($d$): $E_d = 2{{\Delta d} \over d} = 2 \times {{0.005 \text{ mm}} \over {0.5 \text{ mm}}} = 2 \times {{5 \times 10^{-3}} \over {5 \times 10^{-1}}} = 2 \times {1 \over 100} = {2 \over 100} = {1 \over 50} = 0.02$

5. Compare the contributions: We find that $E_l = 0.02$ and $E_d = 0.02$. Therefore: $E_l = E_d$ The contributions to the maximum probable error of the Y measurement due to the errors in the measurements of $d$ and $l$ are the same.

6. Conclusion: Based on the calculation, the correct option is A.