Efficiency of the carnot engine is given by,

$\eta =1-\frac{T_{\sin k}}{T_{source}}$.

Case 1:

$$\begin{gathered} \frac{25}{100}=1-\frac{250}{T_{source}} \\ \Rightarrow \frac{{\displaystyle 250}}{{\displaystyle T_{source}}}=1-\frac{1}{4} \\ \Rightarrow T_{source}=\frac{1000}{3} \mathrm{K} \end{gathered}$$

Case 2:

$$\begin{gathered} \frac{50}{100}=1-\frac{250}{T_{source}} \\ \Rightarrow \frac{{\displaystyle 250}}{{\displaystyle T_{source}}}=1-\frac{1}{2} \\ \Rightarrow T_{source}=500 \mathrm{K} \end{gathered}$$

The temperature of the sink should be increased by, 

$$\begin{gathered} =500 \mathrm{K}-\frac{1000}{3} \mathrm{K} \\ =\frac{1000}{6} \mathrm{K} \end{gathered}$$