The near point and the far point of a child are at 10 cm and 100 cm...

Using the lens formula, the power of the eye-lens is given by $P = \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, where $v$ and $u$ are in meters.

The image distance is fixed at the retina, so $v = +2.0 \text{ cm} = +0.02 \text{ m}$.

For the far point, the object distance is $u = -100 \text{ cm} = -1.0 \text{ m}$.

The minimum power of the eye-lens is:

P_{\text\min} = \dfrac{1}{0.02} - \dfrac{1}{-1.0} = 50 + 1 = +51 \text{ D}

For the near point, the object distance is $u = -10 \text{ cm} = -0.1 \text{ m}$.

The maximum power of the eye-lens is:

P_{\text\max} = \dfrac{1}{0.02} - \dfrac{1}{-0.1} = 50 + 10 = +60 \text{ D}

The range of the power of the eye-lens is $+51 \text{ D}$ to $+60 \text{ D}$.