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Given:
Initial frequency, $f_1 = 4.69 \times 10^{14} \text{ Hz}$
Initial fringe width, $\beta_1 = 2.4 \times 10^{-4} \text{ m}$
New frequency, $f_2 = 7.5 \times 10^{14} \text{ Hz}$

Fringe width in Young's double-slit experiment is given by:
$\beta = \frac{\lambda D}{d} = \frac{c D}{f d}$
Since $c$, $D$, and $d$ are constant:
$\beta \propto \frac{1}{f}$

Therefore,
$\frac{\beta_2}{\beta_1} = \frac{f_1}{f_2}$
$\beta_2 = \beta_1 \times \frac{f_1}{f_2} = 2.4 \times 10^{-4} \times \frac{4.69 \times 10^{14}}{7.5 \times 10^{14}} \approx 1.5008 \times 10^{-4} \text{ m}$

The change in fringe width is:
$\Delta \beta = \beta_1 - \beta_2 = 2.4 \times 10^{-4} \text{ m} - 1.5008 \times 10^{-4} \text{ m} = 8.992 \times 10^{-5} \text{ m} \approx 9.0 \times 10^{-5} \text{ m}$

Hence, the fringe width decreases by $8.992 \times 10^{-5} \text{ m}$ (or $0.08992 \text{ mm}$).