A reversible Carnot heat engine converts $\frac{1}{4}th$  of its input heat into work. When the temperature of the sink is reduced by 50 K, its efficiency becomes $33\frac{1}{3}\%$ . The initial temperatures of the source and the sink respectively are 

Given work,  $W=\frac{Q}{4}$
Where,  $Q=$ input heat.
Efficiency,  $\eta =\frac{W}{Q}=\frac{Q/4}{Q}\Rightarrow \eta =\frac{1}{4}$
Also,  $\eta =1-\frac{T_2}{T_1}$
Where, $T_2=$ temperature of sink
And $T_1=$ temperature of source. 
$1-\frac{T_2}{T_1}=\frac{1}{4}\to \left(1\right)$
When temperature of sink is reduced by 50 K, 
efficiency becomes  $\frac{100}{3}\%$
$\eta =\frac{100}{3}\times \frac{1}{100}=\frac{1}{3}$
So,  $1-\frac{T_2-50}{T_1}=\frac{1}{3}$
$1-\frac{T_2}{T_1}+\frac{50}{T_1}=\frac{1}{3}\to \left(2\right)$
Subtracting Eq. (2) from Eq. (1), we get 
$\frac{50}{T_1}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$
$T_1=600K$
$T_2=\frac{3}{4}\times 600=450K$