\(\omega_1=\frac{\lambda_1 D}{d} \& \omega_2=\frac{\lambda_2 D}{d} \)
\(\omega_1=16 \mathrm{~mm} \& \omega_2=12 \mathrm{~mm} \)
\(\text { so LCM }\left(\omega_1, \omega_2\right)=48 \mathrm{~mm}\)
so at 48 mm distance both bright fringes will be found.