Consider the following floating point format Mantissa is a pure fraction in…

Key idea: store the exponent in excess-64 and encode the mantissa as an 8-bit pure fraction (mantissa/256); do not normalize or round — truncate the fractional part.

1. Write the value as mantissa × 2^exponent. The given number is 0.239 × 2^13, so the true exponent is 13 and the fractional part to encode is 0.239.

2. Encode the mantissa as an 8-bit pure fraction: multiply 0.239 by 256 to get 0.239 × 256 = 61.184. Without rounding, truncate to 61 (decimal) = 0x3D. This gives mantissa = 61/256 ≈ 0.23828125.

3. Encode the exponent using excess-64: stored exponent = 13 + 64 = 77 (decimal) = 0x4D.

4. Assemble the 16-bit word: sign bit = 0 (positive), stored exponent = 0x4D (placed in bits 14..8), mantissa byte = 0x3D (bits 7..0). The final hexadecimal bytes (high-order first) are 0x4D 0x3D.

Final answer: 4D 3D