A normal eye has its retina positioned 2 cm behind the eye-lens....

The image distance for the eye-lens is the distance to the retina, so $v = 2 \text{ cm} = +0.02 \text{ m}$.

When the eye is most strained, it focuses on an object placed at the near point. For a normal eye, the near point is at $25 \text{ cm}$.

The object distance is $u = -25 \text{ cm} = -0.25 \text{ m}$.

Using the lens formula, the power $P$ of the eye-lens is given by:

$P = \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$P = \dfrac{1}{0.02} - \dfrac{1}{-0.25}$

$P = 50 + 4 = 54 \text{ D}$