A vertically projected body attains the maximum height in 6s. The ratio of kinetic energy at the end of 3rd second to decrease in kinetic energy in the next three seconds is

Let H be the maximum height reached in 6 sec

$u=gt=60m/s$

after 3 sec,

$v=u-gt =60-30=30m/s$

Thus,

$\frac{KE\hspace{0.17em}at\hspace{0.17em}\hspace{0.17em}end\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}3rds}{decrease\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}KE\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}next\hspace{0.17em}\hspace{0.17em}3s}=\frac{{\displaystyle \frac{1}{2}}m{\left(30\right)}^2}{{\displaystyle \frac{1}{2}}m{\left(30\right)}^2}=1$

$\frac{KE\hspace{0.17em}after 3sec}{Decrease\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}KE\hspace{0.17em}after\hspace{0.17em}\hspace{0.17em}3sec}=\frac{KE\hspace{0.17em}after\hspace{0.17em}3sec}{PE\hspace{0.17em}after\hspace{0.17em}3sec}=\frac{{\displaystyle \frac{H}{2}}}{{\displaystyle \frac{H}{2}}}=1:1$

Hence the correct answer is 1:1.