A person has a near point at 100 cm . What power of lens is...

The object is to be placed at a distance of $20 \text{ cm}$ from the eye.

Since the spectacles are $2.0 \text{ cm}$ away from the eyes, the object distance from the lens is:
$u = -(20 - 2.0) = -18 \text{ cm}$

The image must be formed at the near point of the person, which is $100 \text{ cm}$ from the eye. The image distance from the lens is:
$v = -(100 - 2.0) = -98 \text{ cm}$

Using the lens formula:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\dfrac{1}{f} = \dfrac{1}{-98} - \dfrac{1}{-18}$

$\dfrac{1}{f} = \dfrac{1}{18} - \dfrac{1}{98}$

The power of the lens $P$ is given by $P = \dfrac{100}{f}$ (where $f$ is in cm):
$P = \dfrac{100}{18} - \dfrac{100}{98}$

$P = 5.55 - 1.02 = +4.53 \text{ D}$