The power of a biconvex lens is 10 dioptre and the radius of...

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

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

$⇒ 10 = \dfrac{10}{2 ( \mu - 1 )}$$⇒ ( \mu - 1 ) = \dfrac{1}{2} ⇒ \mu = \dfrac{1}{2} + 1 = \dfrac{3}{2}$.