Firstly find out distance of retina from his eye lens i.e., image distance in first case (we are assuming distance between eye lens and retina is not changing which is not true practically we are taking that liberty to make problem more mathematically convenient. However it doesn't effect base line of problem)

$\frac{1}{\text{v}}-\frac{1}{-25}=\frac{1}{f_1}$

$\frac{1}{\text{v}}+\frac{1}{25}=\frac{1}{f_1}$

for second case image distance should be same

$\frac{1}{\text{v}}+\frac{1}{50}=\frac{1}{f_2}$

$\Rightarrow \frac{1}{25}-\frac{1}{50}=\frac{1}{f_1}-\frac{1}{f_2}$

$\Rightarrow \frac{1{cm}^{-1}}{50}=\frac{1}{f_1}-\frac{1}{f_2}={\text{P}}_1-{\text{P}}_2=\mathrm{2 } \text{D}$

Let power of his eye lens in two cases are (Pe)1, (Pe)2 respectively

${\text{P}}_1=\left(\text{2.5}+{\left({\text{P}}_{\text{e}}\right)}_1\right)$

${\text{P}}_2=\left(\text{2.5}+{\left({\text{P}}_{\text{e}}\right)}_2\right)$

$\Rightarrow {\text{P}}_1-{\text{P}}_2={\left({\text{P}}_{\text{e}}\right)}_1-{\left({\text{P}}_{\text{e}}\right)}_2$

                  = 2 D

So loss in power is 2 D so he should use additional power of 2 D i.e. 4.5 D