A Carnot engine whose heat sinks at 27 ° C , has an efficiency of...

Efficiency of carnot engine is given by, $η = 1 - \dfrac{{T}_{sink}}{{T}_{source}}$.

For initial case: ${T}_{sink} = \left(27 + 273\right) ° C = 300K$

So we can write, $\dfrac{1}{4} = 1 - \dfrac{300}{{T}_{1}}$$⇒ {T}_{1} = 400K$

Now for the second case: ${T}_{sink} = 300K$

After increasing the previous efficiency by 100%, value of efficiency will get doubled.

Therefore, $\dfrac{1}{2} = 1 - \dfrac{300}{{T}_{2}}$ $⇒ {T}_{2} = 600K$

Increase in temperature required will be, $600K - 400K = 200K$