The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record. Now, the frictional force μR where R is the normal reaction,

and R = mg
Hence the force of friction = μmg and centripetal force required is mυ²/r  or mrω² are the same for both the coins and we have different values of r for the two coins. So to prevent slipping i.e. causing coins to rotate,$\mu mg\ge \mu r{\omega }^2$
$Or, mg\ge r{\omega }^2$

$For first coin:$
$r=4 cm =\frac{4}{100}m$
$n=33\frac{1}{2} rev/min=\frac{100}{3\times 60} rev/s$
$\omega =2\pi n=2\pi n=2\pi \times \frac{100}{180}=3.49 s^{-1}$
$\Rightarrow r{\omega }^2=\frac{4}{100}\times {\left(3.49\right)}^2=0.49 m{s}^{-2}$
$And, \mu g=0.15\times 10=1.5 m{s}^{-2}$
$as \mu g>r{\omega }^2$
$The coin will revolve with record.$
$For second coin:$
$r=14 cm =\frac{4}{100} m$
$\omega =3.49 s^{-1}$
$\Rightarrow r{\omega }^2=\frac{14}{100}\times {\left(3.49\right)}^2$
$=1.705 m{s}^{-2}$
$And, \mu g=1.5 m{s}^{-2}$
$as \mu g<r{\omega }^2$
$The coin will not revolve with record$