Given that the radius of the Earth is decreased to three-fourths of its current value while its mass remains unchanged, we need to determine the new duration of the Earth's day.

First, using the principle of conservation of angular momentum, we have:

$ \tau_{\text{ext}} = 0 \implies \text{Angular momentum is conserved} $

Therefore,

$ \frac{2}{5} M R^2 \cdot \omega_i = \frac{2}{5} M \left(\frac{3R}{4}\right)^2 \cdot \omega_f $

Simplifying this equation, we find:

$ \omega_f = \frac{16}{9} \omega $

Since the period $ T $ of rotation is given by:

$ T = \frac{2\pi}{\omega} $

For the new period $ T_1 $, we have:

$ T_1 = \frac{2\pi}{\omega_f} = \frac{2\pi}{\frac{16}{9}\omega} = \frac{9}{16} \times T $

Given that the initial period $ T $ is 24 hours:

$ T_1 = \frac{9}{16} \times 24 \text{ hours} $

$ T_1 = 13 \text{ hours} ~30 \text{ minutes} $

Thus, the new duration of the Earth's day would be 13 hours and 30 minutes.