Step 1: Concept of magnetic dipole work.
A current-carrying coil behaves like a magnetic dipole having magnetic moment:
\[
m = N I A
\]
Work done in rotating a dipole from \(0^\circ\) to \(180^\circ\) is:
\[
W = 2mB
\]
Step 2: Convert area into SI units.
\[
A = 2 \times 1.25 = 2.5 \, \text{cm}^2 = 2.5 \times 10^{-4} \, \text{m}^2
\]
Step 3: Calculate magnetic moment.
\[
m = 250 \times 55 \times 10^{-6} \times 2.5 \times 10^{-4}
\]
Step 4: Simplify step-by-step.
First multiply current and area:
\[
55 \times 10^{-6} \times 2.5 \times 10^{-4}
= 137.5 \times 10^{-10}
= 1.375 \times 10^{-8}
\]
Now multiply by turns:
\[
m = 250 \times 1.375 \times 10^{-8}
= 3.4375 \times 10^{-6} \, \text{A m}^2
\]
Step 5: Work done for 180° rotation.
\[
W = 2mB
\]
\[
W = 2 \times 3.4375 \times 10^{-6} \times 0.64
\]
\[
W = 4.4 \times 10^{-6} \, \text{J}
\]
Step 6: Final interpretation.
\[
\boxed{4.4 \, \mu \text{J}}
\]