Let ten's digit be x and unit's digit be x2.

Original number = \( \Large 10x + x^{2} \)

New number = \( \Large 10x^{2} + x \)

According to the question,

=\( \Large 10x^{2} + x - 10x - x^{2} = 54 \)

= \( \Large 9x^{2} - 9x = 54 \)

= \( \Large 9 \left(x^{2} - x\right) = 54 \)

= \( \Large x^{2} - x - 6 = 0 \)

= \( \Large x^{2} - 3x + 2x - 6 = 0 \)

= \( \Large x \left(x - 3\right) + 2 \left(x - 3\right) = 0 \)

= \( \Large \left(x - 3\right) \left(x + 2\right) = 0 \)

Therefore, x = 3, -2

Therefore, Ten's digit = x = 3

Unit's digit = \( \Large x^{2} = 3^{2} = 9 \)

Original number = 39

Required number = \( \Large 39 \times \frac{40}{100} = 15.6 \)