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Improved: 16 × 2 = 32 bytes
But ENIAC occupies 200 bytes
Yet the difference is not required.
Re-read: how many total bytes does the ENIAC register memory occupy — pure count from specs:
digits × 10 registers = 200 digit-equivalents, but memory stored in bytes.
Assume 1 digit = 1/8 byte → 20 digits = 20/8 = 2.5 bytes per register
× 2.5 = <<10*2.5=25>>25 bytes
But question asks only for ENIAC: 25
But earlier step said improved to 16 binary registers (2 bytes each) → so 32
But ENIAC: 25