The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm . Then the…

Recall the proper relation between power and focal length, $P=\frac{100}{f}\Rightarrow f=\frac{100}{P}=\frac{100}{10}=10 \mathrm{cm}$, now the relation between focal length and radius of curvature and refractive index, of the spherical surface lens, 

$f=\frac{R}{2(\mu -1)}$ (for equiconvex lens)

$\Rightarrow 10=\frac{10}{2(\mu -1)}$$\Rightarrow (\mu -1)=\frac{1}{2}\Rightarrow \mu =\frac{1}{2}+1=\frac{3}{2}$.