we have to use MAINLY three CONCEPTS which are  L2=L1(1+(alpha)(change in temprature)) another is binomial THEOREM which is (1 + x )n= 1 + nx         for x and the last one which is T = 2\pi\sqrt{\frac{l}{g}} now it is given that clock is 5 seconds fast each day i.e. 1.16 \times 10-4 seconds i.e. time period at that temprature is (2 – 1.16 \times 10-4) and let at \theta temprature it shows correct time i.e. it’s time period is 2 seconds  now, \frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{1}}{L_{o}}} and L1 = Lo(1 + \alpha(15  – \theta)) \Rightarrow \frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{o}(1 + \alpha (15 - \theta ))}{L_{o}}} using binomial theorem \Rightarrow \frac{T_{1}}{T_{o}}=1 - \frac{1}{2}\alpha(15 - \theta ) \frac{2 - 1.16 \times 10^{-4}}{2} = 1 + \frac{1}{2}\alpha (\theta - 15) \theta - 15 = \frac{1.16 \times 10^{-4}}{\alpha}        …......................1 similarly  30 - \theta = \frac{2.32 \times 10^{-4}}{\alpha}   …............................2 now SUBTRACTING adding equation 1 and 2 we get 15 = \frac{3.48 \times 10^{-4}}{\alpha} \Rightarrow \alpha = \frac{3.48 \times 10^{-4}}{15} substituting it in equation 1 we get  \theta - 15 = 5 \Rightarrow \theta = 20 so clock will give correct time at 20 oC