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 mR where R is the normal reaction, and $R = mg$ Hence force of friction = mmg and centripetal force required is $\frac{(mv)^2}{r}$ or $mr\omega^2\mu\omega$ are 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$\mu m g \geq \mu r \omega^{2}$ or $\mu g \geq r \omega^{2}$ ...(i)For 1$^{\text{st}}$ coin $r=4\,\text{cm}= \frac{4}{100}\,\text{m},$$v=33\frac{1}{3}\,\text{rev./min}.$$=\frac{100}{3\times 60}\,\text{rev./s}$$\omega=2\pi v=2\pi\times\frac{100}{180}=3.49\,\text{s}^{-1}$$\therefore r\omega^{2}= \frac{4}{100}\times(3.49)^{2}=0.49\,\text{m}\,\text{s}^{-2}$ and $\mu g=0.15\times 10=1.5\,\text{m}\,\text{s}^{-2}$as $\mu g > r\omega^{2}$, therefore this coin will revolve with record.For 2$^{\text{nd}}$ coin $r=14\,\text{cm}= \frac{4}{100}\,\text{m}, \omega=3.49\,\text{s}^{-1}$$r\omega^{2}= \frac{14}{100}\times(3.49)^{2}=1.705\,\text{m}\,\text{s}^{-2}$ and $\mu g=1.5\,\text{m}\,\text{s}^{-2}$Here, $\mu g \geq r\omega^{2}$ is not satisfied, so this coin will not revolve with record.Note : We have nothing to do with the radius of the record (= 15 cm).